Trigonometrical Identities

Given sin A = $\frac{3}{5}$

sin² A + cos² A = 1

Putting values we get,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{5}\Big)^2 + \text{cos}^2A = 1 \\[1em] \Rightarrow \frac{9}{25} + \text{cos}^2A = 1 \\[1em] \Rightarrow \text{cos}^2A = 1 - \frac{9}{25} \\[1em] \Rightarrow \text{cos}^2A = \frac{25 - 9}{25} \\[1em] \Rightarrow \text{cos}^2 A = \frac{16}{25} \\[1em] \Rightarrow \text{cos} A = \sqrt{\frac{16}{25}} \\[1em] \Rightarrow \text{cos }A = \frac{4}{5}.\end{aligned}$$

sec A = $\frac{1}{\text{cos A}} = \frac{1}{\frac{4}{5}} = \frac{5}{4}$.

cosec A = $\frac{1}{\text{sin A}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}.$

1 + tan² A = sec² A

Putting values we get,

$$\begin{aligned}1 + \text{tan}^2A = \Big(\frac{5}{4}\Big)^2 \\[1em] 1 + \text{tan}^2A = \frac{25}{16} \\[1em] \text{tan}^2A = \frac{25}{16} - 1 \\[1em] \text{tan}^2A = \frac{25 - 16}{16} \\[1em] \text{tan}^2A = \frac{9}{16} \\[1em] \text{tan }A = \sqrt{\frac{9}{16}} \\[1em] \text{tan }A = \frac{3}{4}.\end{aligned}$$

1 + cot² A = cosec² A

Putting values we get,

$$\begin{aligned}1 + \text{cot}^2A = \Big(\frac{5}{3}\Big)^2 \\[1em] 1 + \text{cot}^2A = \frac{25}{9} \\[1em] \text{cot}^2A = \frac{25}{9} - 1 \\[1em] \text{cot}^2A = \frac{25 - 9}{9} \\[1em] \text{cot}^2A = \frac{16}{9} \\[1em] \text{cot }A = \sqrt{\frac{16}{9}} \\[1em] \text{cot }A = \frac{4}{3}.\end{aligned}$$

Hence, the value of,

$$\begin{aligned}\text{cos A} = \frac{4}{5} \\[1em] \text{tan A} = \frac{3}{4} \\[1em] \text{cot A} = \frac{4}{3} \\[1em] \text{sec A} = \frac{5}{4} \\[1em] \text{cosec A} = \frac{5}{3}.\end{aligned}$$

Given sec A = $\frac{17}{8}$

⇒ cos A = $\frac{1}{\text{sec A}} = \frac{1}{\frac{17}{8}} = \frac{8}{17}$.

sin² A + cos² A = 1

Putting values we get,

$$\begin{aligned}\Rightarrow \text{sin}^2A + \Big(\frac{8}{17}\Big)^2 = 1 \\[1em] \Rightarrow \text{sin}^2A + \frac{64}{289} = 1 \\[1em] \Rightarrow \text{sin}^2A = 1 - \frac{64}{289} \\[1em] \Rightarrow \text{sin}^2A = \frac{289 - 64}{289} \\[1em] \Rightarrow \text{sin}^2 A = \frac{225}{289} \\[1em] \Rightarrow \text{sin } A = \sqrt{\frac{225}{289}} \\[1em] \Rightarrow \text{sin }A = \frac{15}{17}.\end{aligned}$$

cosec A = $\frac{1}{\text{sin A}} = \frac{1}{\frac{15}{17}} = \frac{17}{15}.$

1 + tan² A = sec² A

Putting values we get,

$$\begin{aligned}1 + \text{tan}^2A = \Big(\frac{17}{8}\Big)^2 \\[1em] 1 + \text{tan}^2A = \frac{289}{64} \\[1em] \text{tan}^2A = \frac{289}{64} - 1 \\[1em] \text{tan}^2A = \frac{289 - 64}{64} \\[1em] \text{tan}^2A = \frac{225}{64} \\[1em] \text{tan }A = \sqrt{\frac{225}{64}} \\[1em] \text{tan }A = \frac{15}{8}.\end{aligned}$$

1 + cot² A = cosec² A

Putting values we get,

$$\begin{aligned}1 + \text{cot}^2A = \Big(\frac{17}{15}\Big)^2 \\[1em] 1 + \text{cot}^2A = \frac{289}{225} \\[1em] \text{cot}^2A = \frac{289}{225} - 1 \\[1em] \text{cot}^2A = \frac{289 - 225}{225} \\[1em] \text{cot}^2A = \frac{64}{225} \\[1em] \text{cot }A = \sqrt{\frac{64}{225}} \\[1em] \text{cot } A = \frac{8}{15}.\end{aligned}$$

Hence, the value of,

$$\begin{aligned}\text{sin A} = \frac{15}{17} \\[1em] \text{cos A} = \frac{8}{17} \\[1em] \text{tan A} = \frac{15}{8} \\[1em] \text{cot A} = \frac{8}{15} \\[1em] \text{cosec A} = \frac{17}{15}.\end{aligned}$$

Given 12 cosec θ = 13

⇒ cosec θ = $\frac{13}{12}$

1 + cot² θ = cosec² θ

Putting values we get,

$$\begin{aligned}1 + \text{cot}^2\spaceθ = \Big(\frac{13}{12}\Big)^2 \\[1em] 1 + \text{cot}^2\spaceθ = \frac{169}{144} \\[1em] \text{cot}^2\spaceθ = \frac{169}{144} - 1 \\[1em] \text{cot}^2\spaceθ = \frac{169 - 144}{144} \\[1em] \text{cot}^2\spaceθ = \frac{25}{144} \\[1em] \text{cot }θ = \sqrt{\frac{25}{144}} \\[1em] \text{cot } θ = \frac{5}{12}.\end{aligned}$$

We need to find the value of $\frac{2\text{ sin θ} - 3\text{ cos θ}}{4\text{ sin θ} - 9\text{ cos θ}}$

Dividing numerator and denominator of above expression by sin θ.

$$\begin{aligned}\Rightarrow \frac{\frac{2\text{ sin θ} - 3\text{ cos θ}}{\text{ sin θ}}}{\frac{4\text{ sin θ} - 9\text{ cos θ}}{\text{sin θ}}} \\[1em] = \frac{2 - 3\text{ cot θ}}{4 - 9\text{ cot θ}} \\[1em] = \frac{2 - 3 \times \frac{5}{12}}{4 - 9 \times \frac{5}{12}} \\[1em] = \frac{2 - \frac{5}{4}}{4 - \frac{15}{4}} \\[1em] = \frac{\frac{8 - 5}{4}}{\frac{16 - 15}{4}} \\[1em] = \frac{\frac{3}{4}}{\frac{1}{4}} \\[1em] = \frac{3 \times 4}{4} \\[1em] = 3.\end{aligned}$$

Hence, the value of the expression is 3.

We need to find the value of

$\text{cos}^2\space26° + \text{cos 64° sin 26°} + \frac{\text{tan 36°}}{\text{cot 54°}}$

The above equation can be written as,

$\text{cos}^2\space26° + \text{cos (90 - 26)° sin 26°} + \frac{\text{tan 36°}}{\text{cot (90 - 36)°}}$

As, cos(90 - θ) = sin θ and cot(90 - θ) = tan θ. Using in above equation we get,

$$\begin{aligned}\Rightarrow \text{cos}^2\space26° + \text{sin 26° sin 26°} + \frac{\text{tan 36°}}{\text{tan 36°}} \\[1em] = \text{cos}^2\space26° + \text{sin}^2\space26° + \frac{\text{tan 36°}}{\text{tan 36°}} \\[1em] = 1 + 1 \quad [\because \text{sin}^2\space A + \text{cos}^2\space A = 1] \\[1em] = 2.\end{aligned}$$

Hence, the value of the expression is 2.

We need to find the value of

$\frac{\text{sec 17°}}{\text{cosec 73°}} + \frac{\text{tan 68°}}{\text{cot 22°}} + \text{cos}^2\space 44° + \text{cos}^2\space 46°$

The above equation can be written as,

$\frac{\text{sec 17°}}{\text{cosec (90 - 17)°}} + \frac{\text{tan 68°}}{\text{cot (90 - 68)°}} + \text{cos}^2\space (90 - 46)° + \text{cos}^2\space 46°$

As, cos(90 - θ) = sin θ, cosec(90 - θ) = sec θ, cot(90 - θ) = tan θ and sin²θ + cos²θ = 1. Using in above equation we get,

$$\begin{aligned}\Rightarrow \frac{\text{sec 17°}}{\text{sec 17°}} + \frac{\text{tan 68°}}{\text{tan 68°}} + \text{sin}^2\space46° + \text{cos}^2\space46° \\[1em] = 1 + 1 + 1 \\[1em] = 3\end{aligned}$$

Hence, the value of the expression is 3.

We need to find the value of

$\frac{\text{sin 65°}}{\text{cos 25°}} + \frac{\text{cos 32°}}{\text{sin 58°}} - \text{sin 28° sec 62°} + \text{cosec}^2 \space 30°$

As sec θ = $\frac{1}{\text{cos θ}}$

The above equation can be written as,

$\frac{\text{sin 65°}}{\text{cos (90 - 65)°}} + \frac{\text{cos (90 - 58)°}}{\text{sin 58°}} - \text{sin 28°} \frac{1}{\text{cos (90 - 28)°}} + \text{cosec}^2 \space 30°$

As, cos(90 - θ) = sin θ and cosec 30° = 2. Using in above equation,

$$\begin{aligned}= \frac{\text{sin 65°}}{\text{sin 65°}} + \frac{\text{sin 58°}}{\text{sin 58°}} - \text{sin 28°} \frac{1}{\text{sin 28°}} + 2^2 \\[1em] = 1 + 1 - 1 + 4 \\[1em] = 6 - 1 \\[1em] = 5.\end{aligned}$$

Hence, the value of the expression is 5.

We need to find the value of

$\frac{\text{sec 29°}}{\text{cosec 61°}} + \text{2 cot 8° cot 17° cot 45° cot 73° cot 82°} - \text{3(sin}^2 38° + \text{sin}^2 52°)$

The above equation can be written as,

$\frac{\text{sec 29°}}{\text{cosec (90 - 29)°}} + \text{2 cot 8° cot 17° cot 45° cot (90 - 17)° cot (90 - 8)°} - \text{3(sin}^2 (90 - 52)° + \text{sin}^2 52°)$

As, sin(90 - θ) = cos θ, cosec(90 - θ) = sec θ, cot(90 - θ) = tan θ, cot θ.tan θ = 1, cot 45° = 1 and sin²θ + cos²θ = 1. Using this in above equation we get,

$$\begin{aligned}\Rightarrow \frac{\text{sec 29°}}{\text{sec 29°}} + \text{2 cot 8° cot 17° cot 45° tan 17° tan 8°} - \text{3(cos}^2 52° + \text{sin}^2 52°) \\[1em] = 1 + \text{2 cot 8° tan 8° cot 45° cot 17° tan 17°} - \text{3(cos}^2 52° + \text{sin}^2 52°) \\[1em] = 1 + 2 \times 1 \times 1 \times 1 - 3(1) \\[1em] = 1 + 2 - 3 \\[1em] = 3 - 3 \\[1em] = 0.\end{aligned}$$

Hence, the value of the above expression is 0.

We need to find the value of

$\frac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°}$

As, sin(90 - θ) = cos θ, cos(90 - θ) = sin θ, tan(90 - θ) = cot θ, sin²θ + cos²θ = 1 and cosec²θ - cot²θ = 1.

The above equation can be written as,

$$\begin{aligned}\Rightarrow\frac{\text{sin 35° cos (90 - 35)° + cos 35° sin (90 - 35)°}}{\text{cosec}^2 10° - \text{tan}^2 (90 - 10)°} \\[1em] = \frac{\text{sin 35° sin 35° + cos 35° cos 35°}}{\text{cosec}^2 10° - \text{cot}^2 10°} \\[1em] = \frac{\text{sin}^2 35° + \text{cos}^2 35°}{\text{cosec}^2 10° - \text{cot}^2 10°} \\[1em] = \frac{1}{1} \\[1em] = 1.\end{aligned}$$

Hence, the value of the above expression is 1.

We need to find the value of:

sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30°

The above equation can be written as,

sin² 34° + sin² (90 - 34)° + 2 tan 18° tan (90 - 18)° - cot² 30°.

As, sin(90 - θ) = cos θ, tan(90 - θ) = cot θ, sin²θ + cos²θ = 1, tan θ.cot θ = 1 and cot 30° = $\sqrt{3}$. Using in above equation we get,

⇒ sin² 34° + cos² 34° + 2 tan 18° cot 18° - $\sqrt{3}^2$

⇒ 1 + 2 - 3 = 0.

Hence, the value of the above expression is 0.

We need to find the value of,

$\Big(\frac{\text{tan 25°}}{\text{cosec 65°}}\Big)^2 + \Big(\frac{\text{cot 25°}}{\text{sec 65°}}\Big)^2 +$ 2 tan 18° tan 45° tan 72°.

The above equation can be written as,

$$\begin{aligned}\Rightarrow \Big[\frac{\text{tan 25°}}{\text{cosec (90 - 25)°}}\Big]^2 + \Big[\frac{\text{cot 25°}}{\text{sec (90 - 25)°}}\Big]^2 + 2 \text{ tan } 18° \times 1 \times \text{tan } (90 - 18)° \\[1em]\end{aligned}$$

As, sec(90° - θ) = cosec θ, tan(90° - θ) = cot θ and cosec(90° - θ) = sec θ. Using in above equation we get,

$$\begin{aligned}\Rightarrow \Big(\frac{\text{tan 25°}}{\text{sec 25°}}\Big)^2 + \Big(\frac{\text{cot 25°}}{\text{cosec 25°}}\Big)^2 + 2 \text{ tan } 18° \times 1 \times \text{cot } 18° \\[1em] \Rightarrow \Bigg(\frac{\frac{\text{sin 25°}}{\text{cos 25°}}}{\frac{1}{\text{cos 25°}}}\Bigg)^2 + \Bigg(\frac{\frac{\text{cos 25°}}{\text{sin 25°}}}{\frac{1}{\text{sin 25°}}}\Bigg)^2 + 2\text{ tan } 18° \times \frac{1}{\text{tan 18°}} \\[1em] \Rightarrow \text{sin}^2 25° + \text{cos}^2 25° + 2 \\[1em] \Rightarrow 1 + 2 = 3.\end{aligned}$$

Hence, the value of the equation is 3.

We need to find the value of,

(cos² 25° + cos² 65°) + cosec θ sec(90° - θ) - cot θ tan (90° - θ).

The above equation can be written as,

⇒ [cos² 25° + cos² (90 - 25)°] + cosec θ cosec θ - cot θ cot θ

⇒ (cos² 25° + sin² 25°) + cosec θ cosec θ - cot θ cot θ

⇒ 1 + cosec² θ - cot² θ

⇒ 1 + 1

⇒ 2.

Hence, the value of the equation is 2.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

The L.H.S of above equation can be written as,

$$\begin{aligned}\Rightarrow \Big(\frac{\text{1}}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}}\Big)(1 - \text{sin A}) \\[1em] \Rightarrow \Big(\frac{\text{1 + sin A}}{\text{cos A}}\Big)\text{(1 - sin A)} \\[1em] \Rightarrow \frac{1 - \text{sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\text{cos A}} \\[1em] \Rightarrow \text{cos A}.\end{aligned}$$

Since, L.H.S. = cos A = R.H.S. hence, proved that (sec A + tan A)(1 - sin A) = cos A.

The L.H.S of above equation can be written as,

⇒ sec² A.(1 - sin² A)

⇒ sec² A.cos ² A

⇒ $\text{sec}^2 A.\frac{1}{\text{sec}^2 A} = 1.$

Since, L.H.S. = 1 = R.H.S. hence, proved that (1 + tan² A)(1 - sin A)(1 + sin A) = 1.

The L.H.S of above equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A.sin A}} \\[1em] \Rightarrow \frac{1}{\text{cos A.sin A}} \\[1em] \Rightarrow \frac{1}{\text{cos A}} \times \frac{1}{\text{sin A}} \\[1em] \Rightarrow \text{sec A. cosec A}.\end{aligned}$$

Since, L.H.S. = sec A.cosec A = R.H.S., hence proved that tan A + cot A = sec A cosec A.

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow (1 - \text{cos A})\Big(1 + \frac{1}{\text{cos A}}\Big) \\[1em] = (1 - \text{cos A})\Big(\frac{1 + \text{cos A}}{\text{cos A}}\Big) \\[1em] = \frac{1 - \text{cos}^2 A}{\text{cos A}} \\[1em] = \frac{\text{sin}^2 A}{\text{cos A}} \\[1em] = \frac{\text{sin A sin A}}{\text{cos A}} \\[1em] = \text{tan A sin A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that (1 - cos A)(1 + sec A) = tan A sin A.

The L.H.S of above equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 A}{\text{sin}^2 A} - \text{cos}^2 A \\[1em] \Rightarrow \frac{\text{cos}^2 A - \text{cos}^2A.\text{sin}^2 A}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{cos}^2 A(1 - \text{sin}^2 A)}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\text{sin}^2 A} \times \text{cos}^2 A \\[1em] \Rightarrow \text{cot}^2 A\text{ cos}^2 A\end{aligned}$$

Since, L.H.S. = cot² A. cos² A = R.H.S., hence proved that cot² A - cos² A = cot² A cos² A.

The L.H.S of above equation can be written as,

$$\begin{aligned}\Rightarrow 1 + \frac{\text{tan }^2 θ}{1 + \text{sec } θ} \\[1em] \Rightarrow 1 + \frac{\text{sec }^2 θ - 1}{1 + \text{sec } θ} \\[1em] \Rightarrow 1 + \frac{(\text{sec } θ - 1)(\text{sec } θ + 1)}{1 + \text{sec } θ} \\[1em] \Rightarrow 1 + \text{sec } θ - 1\\[1em] \Rightarrow \text{sec } θ.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved 1 + $\frac{\text{tan }^2 θ}{1 + \text{sec } θ}$ = sec θ .

The L.H.S of above equation can be written as,

$$\begin{aligned}\Rightarrow \frac{1 + \frac{1}{\text{cos A}}}{\frac{1}{\text{cos A}}} \\[1em] \Rightarrow \frac{\frac{\text{cos A} + 1}{\text{cos A}}}{\frac{1}{\text{cos A}}} \\[1em] \Rightarrow \frac{(\text{cos A} + 1)\text{ cos A}}{\text{ cos A}} \\[1em] \Rightarrow \text{cos A} + 1 \\[1em] \Rightarrow (1 + \text{cos A}) \times \frac{1 - \text{cos A}}{1 - \text{cos A}} \\[1em] \Rightarrow \frac{1 - \text{cos}^2 A}{1 - \text{cos} A} \\[1em] \Rightarrow \frac{\text{sin}^2 A}{1 - \text{cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{1 + \text{sec A}}{\text{sec A}} = \frac{\text{sin}^2 A}{1 - \text{cos A}}$.

The R.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin A}} + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{1 + cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{1 + cos A}}{\text{sin A}} \times \frac{\text{1 - cos A}}{\text{1 - cos A}} \\[1em] \Rightarrow \frac{\text{1 - cos}^2 \text{A}}{\text{sin A}(\text{1 - cos A})} \\[1em] \Rightarrow \frac{\text{sin}^2 \text{A}}{\text{sin A}(\text{1 - cos A})} \quad [\because \text{sin}^2 \text{A} + \text{cos}^2 \text{A} = 1] \\[1em] \Rightarrow \frac{\text{sin A}}{\text{1 - cos A}}\end{aligned}$$

Since, RHS = LHS, hence proved that $\frac{\text{sin A}}{\text{1 - cos A}} = \text{cosec A} + \text{cot A}$

The L.H.S of above equation can be written as,

$$\begin{aligned}\frac{\text{sin A}}{\text{1 + cos A}} \times \frac{1 - \text{cos A}}{1 - \text{cos A }} \\[1em] \frac{\text{sin A}(1 - \text{cos A})}{(1 + \text{cos A})(1 - \text{cos A})} \\[1em] \frac{\text{sin A}(1 - \text{cos A})}{1 - \text{cos}^2 A} \\[1em] \frac{\text{sin A}(1 - \text{cos A})}{\text{sin}^2 A} \\[1em] \frac{1 - \text{cos A}}{\text{sin} A}.\end{aligned}$$

Since, L.H.S. = R.H.S., hence proved that $\frac{\text{sin A}}{\text{1 + cos A}} = \frac{\text{1 - cos A}}{\text{sin A}}$.

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{1 - \frac{\text{sin}^2 A}{\text{cos}^2 A}}{\frac{\text{cos}^2 A}{\text{sin}^2 A} - 1} \\[1em] \Rightarrow \frac{\frac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos}^2 A}}{\frac{\text{cos}^2 A - \text{sin}^2 A}{\text{sin}^2 A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos}^2 A - \text{sin}^2 A} \times \frac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow 1 \times \text{tan}^2 A \\[1em] \Rightarrow \text{tan}^2 A.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{1 - \text{tan}^2 A}{\text{cot}^2 A - 1} = \text{tan}^2 A$.

The R.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin A}} - \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{1 - cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{1 - cos A}}{\text{sin A}} \times \frac{\text{1 + cos A}}{\text{1 + cos A}} \\[1em] \Rightarrow \frac{1 - \text{cos}^2 A}{\text{sin A(1 + cos A)}} \\[1em] \Rightarrow \frac{\text{sin}^2 A}{\text{sin A(1 + cos A)}} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{1 + cos A}}.\end{aligned}$$

Since, R.H.S. = L.H.S. hence, proved that $\frac{\text{sin A}}{\text{1 + cos A}} = \text{cosec A - cot A}$.

The L.H.S of above equation can be written as,

$$\begin{aligned}\Rightarrow \Big(\frac{1 - \text{tan } θ}{1 - \text{cot } θ}\Big)^2 \\[1em] \Rightarrow \Big(\frac{1 - \text{tan } θ}{1 - \frac{1}{\text{tan } θ}}\Big)^2 \\[1em] \Rightarrow \Big(\frac{(1 - \text{tan } θ). \text{tan } θ}{\text{tan } θ - 1} \Big)^2 \\[1em] \Rightarrow \Big(\frac{-(\text{tan } θ - 1). \text{tan } θ}{\text{tan } θ - 1} \Big)^2 \\[1em] \Rightarrow (-\text{tan } θ)^2 \\[1em] \Rightarrow \text{tan }^2 θ.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved $\Big(\frac{1 - \text{tan } θ}{1 - \text{cot } θ}\Big)^2$ = tan² θ

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{{\frac{1}{\text{cos A}} - 1}}{\frac{1}{\text{cos A}} + 1} \\[1em] \Rightarrow \frac{\frac{1 - \text{cos A}}{\text{cos A}}} {\frac{\text{1 + cos A}}{\text{cos A}}} \\[1em] \Rightarrow \frac{\text{(1 - cos A)} \times \text{cos A}}{\text{(1 + cos A)} \times \text{cos A}} \\[1em] \Rightarrow \frac{\text{1 - cos A}}{\text{1 + cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{sec A - 1}}{\text{sec A + 1}} = \frac{\text{1 - cos A}}{\text{1 + cos A}}$.

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\left(\frac{\sin\theta}{\cos\theta}\right)^2}{\left(\frac{1}{\cos\theta}-1\right)^2} \\[1em] \Rightarrow \frac{\frac{\text{sin}^2 θ}{\text{cos}^2 θ}}{\left(\frac{1-\cos\theta}{\cos\theta}\right)^2} \\[1em]\end{aligned}$$

Using $(\sin^2\theta=1-\cos^2\theta):$

$$\begin{aligned}\Rightarrow \frac{\frac{1-\cos^2\theta}{\cos^2\theta}}{\frac{(1-\cos\theta)^2}{\cos^2\theta}} \\[1em] \Rightarrow \frac{1-\cos^2\theta}{(1-\cos\theta)^2} \\[1em] \Rightarrow \frac{(1-\cos\theta)(1+\cos\theta)}{(1-\cos\theta)^2} \\[1em] \Rightarrow \frac{\cancel{(1-\cos\theta)}(1+\cos\theta)}{(1-\cos\theta)^{\cancel{2}}} \\[1em] \Rightarrow \frac{1 + \cos\theta}{1 - \cos\theta} \\[1em]\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{tan}^2 θ}{\text{(sec θ - 1)}^2} = \frac{\text{1 + cos θ}}{\text{1 - cos θ}}$.

The L.H.S of the equation can be written as,

⇒ 1 + tan² A + 2 tan A + 1 + tan² A - 2 tan A

⇒ 2(1 + tan² A)

⇒ 2 sec² A.

Since, L.H.S. = R.H.S hence, proved that (1 + tan A)² + (1 - tan A)² = 2 sec² A.

The L.H.S of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{1}{\text{cos}^2 A} + \frac{1}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{sin}^2 A +\text{cos}^2 A}{\text{cos}^2 A .\text{ sin}^2 A} \\[1em] \Rightarrow \frac{1}{\text{cos}^2 A .\text{ sin}^2 A} \\[1em] \Rightarrow \text{sec}^2 A. \text{cosec}^2 A\end{aligned}$$

Since, L.H.S. = R.H.S hence, proved that sec² A + cosec² A = sec² A cosec² A.

The L.H.S of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{(1 + \text{sin A})^2 + \text{cos}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{1 + \text{sin}^2 A + 2\text{sin A} + \text{cos}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{1 + \text{sin}^2 A + \text{cos}^2 A + 2\text{sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{1 + 1 + 2\text{sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{2 + 2\text{sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{2(1 + \text{sin A})}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{2}{\text{cos A}} \\[1em] \Rightarrow 2\text{ sec A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that $\frac{\text{1 + sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{1 + sin A}} = 2\text{ sec A}$.

The L.H.S of the equation can be written as,

$$\begin{aligned}\Rightarrow \text{tan A}\Big(\frac{1}{\text{sec A - 1}} + \frac{1}{\text{sec A + 1}}\Big) \\[1em] \Rightarrow \text{tan A}\Big[\frac{\text{sec A + 1 + sec A - 1}}{\text{(sec A - 1)(sec A + 1)}}\Big] \\[1em] \Rightarrow \text{tan A}\Big(\frac{\text{2 sec A}}{\text{sec}^2 A - 1}\Big) \\[1em] \Rightarrow \text{tan A}\Big(\frac{\text{2 sec A}}{\text{tan}^2 A}\Big) \\[1em] \Rightarrow \frac{\text{2 sec A}}{\text{tan A}} \\[1em] \Rightarrow \frac{2\frac{1}{\text{cos A}}}{\frac{\text{sin A}}{\text{cos A}}} \\[1em] \Rightarrow \frac{2}{\text{sin A}} \\[1em] \Rightarrow \text{2 cosec A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{tan A}}{\text{sec A - 1}} + \frac{\text{tan A}}{\text{sec A + 1}} = 2 \text{ cosec A}$.

The L.H.S of the equation can be written as,

$$\begin{aligned}\Rightarrow \text{cosec A}\Big(\frac{1}{\text{cosec A - 1}} + \frac{1}{\text{cosec A + 1}}\Big) \\[1em] \Rightarrow \text{cosec A}\Big[\frac{\text{cosec A + 1 + cosec A - 1}}{\text{(cosec A - 1)(cosec A + 1)}}\Big] \\[1em] \Rightarrow \text{cosec A}\Big(\frac{\text{2 cosec A}}{\text{cosec}^2 A - 1}\Big) \\[1em] \Rightarrow \text{cosec A}\Big(\frac{\text{2 cosec A}}{\text{cot}^2 A}\Big) \\[1em] \Rightarrow \frac{\text{2 cosec}^2 A}{\text{cot}^2 A} \\[1em] \Rightarrow \frac{2\frac{1}{\text{sin}^2 A}}{\frac{\text{cos}^2 A}{\text{sin}^2 A}} \\[1em] \Rightarrow \frac{2}{\text{cos}^2 A} \\[1em] \Rightarrow \text{2 sec}^2 A.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{cosec A}}{\text{cosec A - 1}} + \frac{\text{cosec A}}{\text{cosec A + 1}} = 2 \text{ sec}^2 A$.

The L.H.S of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{cos A}}{\text{sin A}} - \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A - \text{sin}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A - \text{(1 - cos}^2 A)}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A + \text{cos}^2 A - 1}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{2\text{cos}^2 A - 1}{\text{sin A cos A}} \\[1em]\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that cot A - tan A = $\frac{2\text{cos}^2 A - 1}{\text{sin A cos A}}$.

The L.H.S of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\frac{\text{cos A}}{\text{sin A}} - 1}{2 - \frac{1}{\text{cos}^2 A}} \\[1em] \Rightarrow \frac{\frac{\text{cos A} - \text{sin A}}{\text{sin A}}}{\frac{2 \text{cos}^2 A - 1}{\text{cos}^2 A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}(2\text{cos}^2 A - 1)} \\[1em] \Rightarrow \frac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}[2\text{cos}^2 A - (\text{sin}^2 A + \text{cos}^2 A)]} \\[1em] \Rightarrow \frac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}(2\text{cos}^2 A - \text{sin}^2 A - \text{cos}^2 A)} \\[1em] \Rightarrow \frac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}(\text{cos}^2 A - \text{sin}^2 A)} \\[1em] \Rightarrow \frac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}(\text{cos A - sin A}) (\text{cos A + sin A})} \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\text{sin A(cos A + sin A)}} \\[1em] \Rightarrow \frac{\text{cos A}}{\text{sin A}} \times \frac{\text{cos A}}{\text{cos A + sin A}} \\[1em] \Rightarrow \frac{\text{cot A. cos A}}{\text{(cos A + sin A)}} \\[1em]\end{aligned}$$

Dividing the numerator and denominator by cos A, we get :

$$\begin{aligned}\Rightarrow \frac{\frac{\text{cot A. cos A}}{\text{cos A}}}{\frac{\text{(cos A + sin A)}}{\text{cos A}}} \\[1em] \Rightarrow \frac{\text{cot A}}{\frac{\text{cos A}}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}}} \\[1em] \Rightarrow \frac{\text{cot A}}{\text{1 + tan A}}\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cot A - 1}}{2 - \text{sec}^2 A} = \frac{\text{cot A}}{\text{1 + tan A}}$.

The L.H.S of above equation can be written as,

$$\begin{aligned}\Rightarrow \frac{1}{1 + \text{sin } θ} + \frac{1}{1 - \text{sin } θ} \\[1em] \Rightarrow \frac{1 \times (1 - \text{sin } θ) + 1 \times (1 + \text{sin } θ)}{(1 + \text{sin } θ) \times (1 - \text{sin } θ)} \\[1em] \Rightarrow \frac{1 - \text{sin } θ + 1 + \text{sin } θ}{1^2 - (\text{sin } θ)^2} \\[1em] \Rightarrow \frac{2}{1 - \text{sin }^2 θ} \\[1em] \Rightarrow \frac{2}{\text{cos }^2 θ} \\[1em] \Rightarrow 2\text{ sec }^2 θ\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved $\frac{1}{1 + \text{sin } θ} + \frac{1}{1 - \text{sin } θ}$ = 2 sec² θ

The L.H.S of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{sin}^2 θ}{\text{cos}^2 θ} - \text{sin}^2 θ \\[1em] \Rightarrow \frac{\text{sin}^2 θ - \text{sin}^2 θ. \text{cos}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow \frac{\text{sin}^2 θ(\text{1 - cos}^2 θ)}{\text{cos}^2 θ} \\[1em] \Rightarrow \text{tan}^2 θ. \text{ sin}^ 2 θ\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that tan² θ - sin² θ = tan² θ sin² θ.

The L.H.S of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{cos θ}}{\text{1 - tan θ}} - \frac{\text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \frac{\text{cos θ}}{{1 - \frac{\text{sin θ}}{\text{cos θ}}}} - \frac{\text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \frac{\text{cos}^2 θ}{\text{cos θ - sin θ}} - \frac{\text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \frac{\text{cos}^2 θ - \text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \frac{\text{(cos θ - sin θ)(cos θ + sin θ)}}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \text{cos θ + sin θ}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{cos θ}}{\text{1 - tan θ}} - \frac{\text{sin}^2 θ}{\text{cos θ - sin θ}} = \text{cos θ + sin θ}$.

The L.H.S of above equation can be written as,

$$\begin{aligned}\Rightarrow \frac{(1 + \text{sin }θ)^2 + (1 - \text{sin }θ)^2}{2\text{cos }^2θ} \\[1em] \Rightarrow\frac{1 + \text{sin }^2θ + 2\text{sin }θ + 1 + \text{sin }^2θ - 2\text{sin }θ}{2\text{cos }^2θ}\\[1em] \Rightarrow\frac{2 + 2\text{sin }^2θ}{2\text{cos }^2θ}\\[1em] \Rightarrow\frac{2(1 + \text{sin }^2θ)}{2\text{cos }^2θ}\\[1em] \Rightarrow\frac{1 + \text{sin }^2θ}{\text{cos }^2θ}\\[1em] \Rightarrow\frac{1}{\text{cos }^2θ} + \frac{\text{sin }^2θ}{\text{cos }^2θ}\\[1em] \Rightarrow \text{sec }θ^2 + \text{tan }^2θ\\[1em]\end{aligned}$$

Since, L.H.S. = sec²θ + tan²θ = R.H.S.

Hence, proved $\frac{(1 + \text{sin }θ)^2 + (1 - \text{sin }θ)^2}{2\text{cos }^2θ}$ = sec²θ + tan²θ.

The L.H.S of the equation can be written as,

⇒ cosec² θ(cosec² θ - 1)
⇒ cosec² θ. cot² θ
⇒ (1 + cot² θ). cot² θ
⇒ cot² θ + cot⁴ θ.

Since, L.H.S. = R.H.S. hence, proved that cosec⁴ θ - cosec² θ = cot⁴ θ + cot² θ.

The L.H.S of the equation can be written as,

⇒ 2(1 + tan² θ) - (sec² θ)² - 2(1 + cot² θ) + (cosec² θ)²
⇒ 2 + 2tan² θ - (1 + tan² θ)² - 2 - 2cot² θ + (1 + cot² θ)²
⇒ 2 + 2tan² θ - (1 + tan⁴ θ + 2tan² θ) - 2 - 2cot² θ + (1 + cot⁴ θ + 2cot² θ)
⇒ 2 + 2tan² θ - 1 - tan⁴ θ - 2tan² θ - 2 - 2cot² θ + 1 + cot⁴ θ + 2cot² θ
⇒ 2 - 2 + 2tan² θ - 2tan² θ + 2cot² θ - 2cot² θ + 1 - 1 + cot⁴ θ - tan⁴ θ
⇒ cot⁴ θ - tan⁴ θ.

Since, L.H.S. = R.H.S. hence, proved that 2 sec² θ - sec⁴ θ - 2 cosec² θ + cosec⁴ θ = cot⁴ θ - tan⁴ θ.

The L.H.S of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{1 + cos θ - (1 - cos}^2 θ)}{\text{sin θ(1 + cos θ)}} \\[1em] \Rightarrow \frac{1 - 1 + \text{cos}^2 θ + \text{cos } θ}{\text{sin θ(1 + cos θ)}} \\[1em] \Rightarrow \frac{\text{cos θ(cos θ + 1)}}{\text{sin θ(1 + cos θ)}} \\[1em] \Rightarrow \frac{\text{cos θ}}{\text{sin θ}} \\[1em] \Rightarrow \text{cot θ}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{1 + cos θ - sin}^2 θ}{\text{sin θ(1 + cos θ)}}$ = cot θ.

As, a³ - b³ = (a - b)(a² + ab + b²)

∴ tan³ θ - (1)³ = (tan θ - 1)(tan² θ + tan θ + 1)

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{(tan θ - 1)}(\text{tan}^2 \text{ θ} + \text{tan θ + 1})}{\text{tan θ - 1}} \\[1em] \Rightarrow \text{tan}^2 \text{ θ} + \text{tan θ + 1} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} - 1 + 1 + \text{tan θ} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} + \text{tan θ}\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{tan}^3 \text{ θ} - 1}{\text{tan θ - 1}}$ = sec² θ + tan θ.

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{1 + cosec A}}{\text{cosec A}} \\[1em] \Rightarrow \frac{1 + \frac{1}{\text{sin A}}}{\frac{1}{\text{sin A}}} \\[1em] \Rightarrow \frac{\frac{\text{sin A + 1}}{\text{sin A}}}{\frac{1}{\text{sin A}}} \\[1em] \Rightarrow \frac{\text{sin A(sin A + 1)}}{\text{sin A}} \\[1em] \Rightarrow \text{sin A + 1}\end{aligned}$$

The R.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{1 - \text{sin}^2 A}{1 - \text{sin A}} \\[1em] \Rightarrow \frac{\text{(1 - sin A)(1 + sin A)}}{\text{1 - sin A}} \\[1em] \Rightarrow \text{1 + sin A}\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{1 + cosec A}}{\text{cosec A}} = \frac{\text{cos}^2 \text{ A}}{\text{1 - sin A}}$

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \sqrt{\frac{(1 - \text{cos A})(1 + \text{cos A})}{\text{(1 + cos A)(1 + \text{cos A})}}} \\[1em] \Rightarrow \sqrt{\frac{(1 - \text{cos}^2 A)}{\text{(1 + cos A)}^2}} \\[1em] \Rightarrow \sqrt{\frac{\text{sin}^2 A}{\text{(1 + cos A)}^2}} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{1 + cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\sqrt{\frac{1 - \text{cos A}}{\text{1 + cos A}}} = \frac{\text{sin A}}{\text{1 + cos A}}$

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \sqrt{\frac{\text{(1 + sin A)(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}}} \\[1em] \Rightarrow \sqrt{\frac{\text{(1 + sin A)}^2}{\text{(1 - sin}^2 A)}} \\[1em] \Rightarrow \sqrt{\frac{\text{(1 + sin A)}^2}{\text{cos}^2 A}} \\[1em] \Rightarrow \frac{\text{1 + sin A}}{\text{cos A}} \\[1em] \Rightarrow \frac{1}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + tan A}\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\sqrt{\frac{\text{1 + sin A}}{\text{1 - sin A}}}$ = tan A + sec A

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \sqrt{\frac{\text{(1 - cos A)(1 - cos A)}}{\text{(1 + cos A)(1 - cos A)}}} \\[1em] \Rightarrow \sqrt{\frac{\text{(1 - cos A)}^2}{\text{(1 - cos}^2 A)}} \\[1em] \Rightarrow \sqrt{\frac{\text{(1 - cos A)}^2}{\text{sin}^2 A}} \\[1em] \Rightarrow \frac{\text{1 - cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{1}{\text{sin A}} - \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A - cot A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\sqrt{\frac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A - cot A.

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\sqrt{(\text{sec A - 1})^2} + \sqrt{(\text{sec A + 1})^2}}{\sqrt{\text{(sec A + 1)(sec A - 1)}}} \\[1em] \Rightarrow \frac{\text{sec A - 1 + sec A + 1}}{\sqrt{\text{sec}^2 A - 1}} \\[1em] \Rightarrow \frac{\text{2 sec A}}{\sqrt{\text{tan}^2 A}} \\[1em] \Rightarrow \frac{\text{2 sec A}}{\text{tan A}} \\[1em] \Rightarrow \frac{\frac{2}{\text{cos A}}}{\frac{\text{sin A}}{\text{cos A}}} \\[1em] \Rightarrow \frac{2}{\text{sin A}} \\[1em] \Rightarrow 2\text{ cosec A}\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\sqrt{\frac{\text{sec A - 1}}{\text{sec A + 1}}} + \sqrt{\frac{\text{sec A + 1}}{\text{sec A - 1}}}$ = 2 cosec A.

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{cos A} \times \frac{\text{cos A}}{\text{sin A}}}{1 - \text{sin A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\text{sin A(1 - sin A)}} \\[1em] \Rightarrow \frac{\text{1 - sin}^2 A}{\text{sin A(1 - sin A)}} \\[1em] \Rightarrow \frac{\text{(1 - sin A)(1 + sin A)}}{\text{sin A(1 - sin A)}} \\[1em] \Rightarrow \frac{\text{1 + sin A}}{\text{sin A}} \\[1em] \Rightarrow \frac{1}{\text{sin A}} + \frac{\text{sin A}}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A + 1}\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{cos A cot A}}{\text{1 - sin A}}$ = 1 + cosec A.

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{1 + \frac{\text{sin A}}{\text{cos A}}}{\text{sin A}} + \frac{1 + \frac{\text{cos A}}{\text{sin A}}}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{cos A}\Big(1 + \frac{\text{sin A}}{\text{cos A}}\Big) + \text{sin A}\Big(1 + \frac{\text{cos A}}{\text{sin A}}\Big)}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{cos A + sin A + sin A + cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{2(cos A + sin A)}}{\text{sin A cos A}} \\[1em] \Rightarrow 2\Big(\frac{\text{cos A}}{\text{sin A cos A}} + \frac{\text{sin A}}{\text{sin A cos A}}\Big) \\[1em] \Rightarrow 2\Big(\frac{1}{\text{sin A}} + \frac{1}{\text{cos A}}\Big) \\[1em] \Rightarrow 2\text{(cosec A + sec A)}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{1 + tan A}}{\text{sin A}} + \frac{\text{1 + cot A}}{\text{cos A}}$ = 2(sec A + cosec A).

The L.H.S of above equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{sin }A}{(1 + \text{cot }A)} - \frac{\text{cos }A}{(1 + \text{tan }A)} \\[1em] \Rightarrow \frac{\text{sin }A}{\Big(1 + \frac{\text{cos }A}{\text{sin }A}\Big)} - \frac{\text{cos }A}{\Big(1 + \frac{\text{sin }A}{\text{cos }A}\Big)} \\[1em] \Rightarrow \frac{\text{sin }A}{\Big(\frac{\text{sin }A + \text{cos }A}{\text{sin }A}\Big)} - \frac{\text{cos }A}{\Big(\frac{\text{cos }A + \text{sin }A}{\text{cos }A} \Big)} \\[1em] \Rightarrow \frac{\text{sin }A \times \text{sin }A}{\text{sin }A + \text{cos }A} - \frac{\text{cos }A \times \text{cos }A}{\text{cos }A + \text{sin }A} \\[1em] \Rightarrow \frac{\text{sin }^2A - \text{cos }^2A}{\text{sin }A + \text{cos }A}\\[1em] \Rightarrow \frac{(\text{sin }A - \text{cos }A)(\text{sin }A + \text{cos }A)}{\text{sin }A + \text{cos }A}\\[1em] \Rightarrow \text{sin }A - \text{cos }A.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved $\frac{\text{sin }A}{(1 + \text{cot }A)} - \frac{\text{cos }A}{(1 + \text{tan }A)}$ = sin A - cos A

The L.H.S. of the equation can be written as,

⇒ (sec² A - tan² A)(sec² A + tan² A)

⇒ 1 × (sec² A + tan² A)

⇒ (sec² A + tan² A)

⇒ (1 + tan² A + tan² A)

⇒ 1 + 2 tan² A

Since, L.H.S. = R.H.S. hence, proved that sec⁴ A - tan⁴ A = 1 + 2tan² A.

a³ - b³ = (a - b)³ + 3ab(a - b)

∴ L.H.S. of the equation can be written as,

⇒ cosec⁶ A - cot⁶ A = (cosec² A - cot² A )³ + 3cosec² A cot² A(cosec² A - cot² A)

⇒ 1³ + 3cosec² A cot² A × 1
⇒ 1 + 3cosec² A cot² A

Since, L.H.S. = R.H.S. hence, proved that cosec⁶ A - cot⁶ A = 3cot² A cosec² A + 1.

a³ - b³ = (a - b)³ + 3ab(a - b)

∴ L.H.S. of the equation can be written as,

⇒ sec⁶ A - tan⁶ A = (sec² A - tan² A)³ + 3sec² A tan² A(sec² A - tan² A)

⇒ 1³ + 3sec² A tan² A × 1

⇒ 1 + 3sec² A tan² A

⇒ 1 + 3(1 + tan² A)tan² A

⇒ 1 + 3(tan⁴ A + tan² A)

⇒ 1 + 3 tan² A + 3 tan⁴ A

Since, L.H.S. = R.H.S. hence, proved that sec⁶ A - tan⁶ A = 1 + 3tan² A + 3tan⁴ A.

L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\frac{\text{cos θ}}{\text{sin θ}} + \frac{1}{\text{sin θ}} - 1}{\frac{\text{cos θ}}{\text{sin θ}} - \frac{1}{\text{sin θ}} + 1} \\[1em] \Rightarrow \frac{\frac{\text{cos θ + 1 - sin θ}}{\text{sin θ}}}{\frac{\text{cos θ - 1 + sin θ}}{\text{sin θ}}} \\[1em] \Rightarrow \frac{{\text{cos θ + 1 - sin θ}}}{\text{cos θ - 1 + sin θ}} \\[1em] \Rightarrow \frac{{\text{cos θ + (1 - sin θ)}}}{\text{cos θ - (1 - sin θ)}} \\[1em] \Rightarrow \frac{{\text{cos θ + (1 - sin θ)}}}{\text{cos θ - (1 - sin θ)}} \times \frac{{\text{cos θ + (1 - sin θ)}}}{{\text{cos θ + (1 - sin θ)}}} \\[1em] \Rightarrow \frac{[\text{cos θ + (1 - sin θ)}]^2}{\text{cos}^2 θ - (\text{1 - sin θ})^2} \\[1em] \Rightarrow \frac{\text{cos}^2 \text{ θ} + \text{(1 - sin θ)}^2 + 2\text{cos θ(1 - sin θ)}}{\text{cos}^2 \text{ θ} - (1 + \text{sin}^2 θ - \text{2sin θ})} \\[1em] \Rightarrow \frac{\text{cos}^2 θ + \text{sin}^2 θ + 1 + \text{2cos θ - 2sin θ - 2 sin θ cos θ}}{\text{cos}^2 \text{ θ} - 1 - \text{sin}^2 θ + \text{2sin θ}} \\[1em] \Rightarrow \frac{1 + 1 + \text{2cos θ - 2sin θ - 2 sin θ cos θ}}{1 - \text{sin}^2 θ - 1 - \text{sin}^2 θ + \text{2 sin θ}} \\[1em] \Rightarrow \frac{2 + \text{2cos θ - 2sin θ - 2sin θ cos θ}}{\text{2sin θ - 2sin}^2 θ} \\[1em] \Rightarrow \frac{\text{2(1 + cos θ) - 2sin θ(1 + cos θ)}}{\text{2sin θ(1 - sin θ)}} \\[1em] \Rightarrow \frac{\text{(1 + cos θ)(2 - 2sin θ)}}{\text{2sin θ(1 - sin θ)}} \\[1em] \Rightarrow \frac{\text{2(1 + cos θ)(1 - sin θ)}}{\text{2sin θ(1 - sin θ)}} \\[1em] \Rightarrow \frac{\text{1 + cos θ}}{\text{sin θ}}.]\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that $\frac{\text{cot θ + cosec θ - 1}}{\text{cot θ - cosec θ + 1}} = \frac{\text{1 + cos θ}}{\text{sin θ}}$.

L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{sin θ}}{\frac{\text{cos θ + 1}}{\text{sin θ}}} \\[1em] \Rightarrow \frac{\text{sin}^2 θ}{\text{1 + cos θ}} \\[1em] \Rightarrow \frac{\text{1 - cos}^2 θ}{\text{1 + cos θ}} \\[1em] \Rightarrow \frac{\text{(1 - cos θ)(1 + cos θ)}}{\text{1 + cos θ}} \\[1em] \Rightarrow \text{1 - cos θ}.\end{aligned}$$

R.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow 2 + \frac{\text{sin θ}}{\frac{\text{cos θ}}{\text{sin θ}} - \frac{1}{\text{sin θ}}} \\[1em] \Rightarrow 2 + \frac{\text{sin}^2 θ}{\text{cos θ - 1}} \\[1em] \Rightarrow \frac{\text{2(cos θ - 1)} + \text{sin}^2 θ}{\text{cos θ - 1}} \\[1em] \Rightarrow \frac{\text{2(cos θ - 1)} + \text{1 - cos}^2 θ}{\text{cos θ - 1}} \\[1em] \Rightarrow \frac{\text{2(cos θ - 1) + (1 + cos θ)(1 - cos θ)}}{\text{cos θ - 1}} \\[1em] \Rightarrow \frac{\text{2(cos θ - 1) - (cos θ - 1)(1 + cos θ)}}{\text{cos θ - 1}} \\[1em] \Rightarrow \frac{\text{(cos θ - 1)[2 - (1 + cos θ)]}}{\text{cos θ - 1}} \\[1em] \Rightarrow \text{2 - 1 - cos θ} \\[1em] \Rightarrow \text{1 - cos θ}\end{aligned}$$

Since, L.H.S. = 1 - cos θ = R.H.S. hence proved that,

$\frac{\text{sin θ}}{\text{cot θ + cosec θ}} = 2 + \frac{\text{sin θ}}{\text{cot θ - cosec θ}}$.

L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \text{(sin θ + cos θ)}\Big(\frac{1}{\text{cos θ}} + \frac{1}{\text{sin θ}}\Big) \\[1em] \Rightarrow \frac{\text{(sin θ + cos θ)(sin θ + cos θ)}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \frac{\text{sin}^2 θ + \text{cos}^2 θ + \text{2 sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \frac{1 + \text{2 sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \frac{1}{\text{sin θ cos θ}} + \frac{\text{2 sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \text{cosec θ sec θ} + 2.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that, (sin θ + cos θ)(sec θ + cosec θ) = 2 + sec θ cosec θ.

L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \Big(\frac{1}{\text{sin A}} - \text{sin A}\Big)\Big(\frac{1}{\text{cos A}} - \text{cos A}\Big) \times \frac{1}{\text{cos}^2 A} \\[1em] \Rightarrow \Big(\frac{\text{1 - sin}^2 A}{\text{sin A}}\Big)\Big(\frac{\text{1 - cos}^2 A}{\text{cos A}}\Big) \times \frac{1}{\text{cos}^2 A} \\[1em]\end{aligned}$$

As 1 - sin² A = cos² A and 1 - cos² A = sin² A.

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 A \text{ sin}^2 A}{\text{sin A cos A}} \times \frac{1}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{cos}^2 A \text{ sin}^2 A}{\text{cos}^3 A \text{ sin} A} \\[1em]\end{aligned}$$

Dividing numerator and denominator by sin A cos A.

$$\begin{aligned}\Rightarrow \frac{\text{sin A cos A}}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{tan A}\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that, (cosec A - sin A)(sec A - cos A)sec² A = tan A.

We know that,

a³ + b³ = (a + b)(a² - ab + b²).

and

a³ - b³ = (a - b)(a² + ab + b²).

Using above formulas, the L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{(sin A + cos A)(sin}^2 A - \text{sin A cos A + cos}^2 A)}{\text{sin A + cos A}} + \frac{\text{(sin A - cos A)(sin}^2 A + \text{sin A cos A + cos}^2 A)}{\text{sin A - cos A}} \\[1em] \Rightarrow \frac{\cancel{\text{(sin A + cos A)}}\text{(sin}^2 A - \text{sin A cos A + cos}^2 A)}{\cancel{\text{(sin A + cos A)}}} + \frac{\cancel{\text{(sin A - cos A)}} \text{(sin}^2 A + \text{sin A cos A + cos}^2 A)}{\cancel{\text{(sin A - cos A)}}} \\[1em] \Rightarrow \text{sin}^2 A + \text{cos}^2 A - \text{sin A cos A} + \text{sin}^2 A + \text{cos}^2 A + \text{sin A cos A} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that, $\frac{\text{sin}^3 A + \text{cos}^3 A}{\text{sin A + cos A}} + \frac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin A - cos A}} = 2$.

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \frac{\frac{1}{\text{tan}^2 A}}{1 + \frac{1}{\text{tan}^2 A}} \\[1em] = \frac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \frac{\frac{1}{\text{tan}^2 A}}{\frac{\text{tan}^2 A + 1}{\text{tan}^2 A}} \\[1em] = \frac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \frac{1}{\text{tan}^2 A + 1} \\[1em] = \frac{\text{tan}^2 A + 1}{\text{tan}^2 A + 1} \\[1em] 1.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that, $\frac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \frac{\text{cot}^2 A}{\text{1 + cot}^2 A} = 1.$

The equation can be written as,

$\frac{1}{\text{sec A + tan A}} + \frac{1}{\text{sec A - tan A}} = \frac{2}{\text{cos A}}$

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{sec A - tan A + sec A + tan A}}{\text{(sec A - tan A)(sec A + tan A)}} \\[1em] = \frac{\text{2 sec A}}{\text{sec}^2 A - \text{tan}^2 A} \\[1em] = \text{2 sec A} \\[1em] = \frac{2}{\text{cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

$\frac{1}{\text{sec A + tan A}} - \frac{1}{\text{cos A}} = \frac{1}{\text{cos A}} - \frac{1}{\text{sec A - tan A}}$.

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \text{sin}^2 A + \text{sec}^2 A + \text{2 sin A sec A} + \text{cos}^2 A + \text{cosec}^2 A + \text{2 cos A cosec A} \\[1em] = \text{sin}^2 A + \text{cos}^2 A + \text{sec}^2 A + \text{cosec}^2 A + \frac{\text{2 sin A}}{\text{cos A}} + \frac{\text{2 cos A}}{\text{sin A}} \\[1em] = 1 + \frac{1}{\text{cos}^2 A} + \frac{1}{\text{sin}^2 A} + \frac{\text{2sin}^2 A + \text{2cos}^2 A}{\text{sin A cos A}} \\[1em] = 1 + \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A \text{ cos}^2 A} + \frac{2(\text{sin}^2 A + \text{cos}^2 A)}{\text{sin A cos A}} \\[1em] = 1 + \frac{1}{\text{sin}^2 A \text{ cos}^2 A} + \frac{2}{{\text{sin A cos A}}} \\[1em] = \Big(1 + \frac{1}{\text{sin A cos A}}\Big)^2 \\[1em] = \Big(\text{1 + sec A cosec A}\Big)^2.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that, (sin A + sec A)² + (cos A + cosec A)² = (1 + sec A cosec A)².

The L.H.S. of the equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\frac{\text{sin A}}{\text{cos A}} + \text{sin A}}{\frac{\text{sin A}}{\text{cos A}} - \text{sin A}} \\[1em] = \frac{\text{sin A}\Big(\frac{1}{\text{cos A}} + 1\Big)}{\text{sin A}\Big(\frac{1}{\text{cos A}} - 1\Big)} \\[1em] = \frac{\Big(\frac{1}{\text{cos A}} + 1\Big)}{\Big(\frac{1}{\text{cos A}} - 1\Big)} \\[1em] = \frac{\text{sec A + 1}}{\text{sec A - 1}}\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that, $\frac{\text{tan A + sin A}}{\text{tan A - sin A}} = \frac{\text{sec A + 1}}{\text{sec A - 1}}.$

Given,

    sin θ + cos θ = $\sqrt{2}$ sin(90° - θ)
⇒ sin θ + cos θ = $\sqrt{2}$ cos θ

Dividing both sides by sin θ

⇒ 1 + cot θ = $\sqrt{2}$ cot θ
⇒ 1 = $\sqrt{2}$ cot θ - cot θ
⇒ 1 = cot θ($\sqrt{2}$ - 1)
⇒ cot θ = $\frac{1}{\sqrt{2} - 1}$

Rationalizing,

$$\begin{aligned}⇒ \text{cot θ} = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} \\[1em] ⇒ \text{cot θ} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1.\end{aligned}$$

Hence, proved that cot θ = $\sqrt{2}$ + 1.

Given,

⇒ 7 sin² θ + 3 cos² θ = 4

⇒ 4 sin² θ + 3 sin² θ + 3 cos² θ = 4

⇒ 4 sin² θ + 3(sin² θ + cos² θ) = 4

⇒ 4 sin² θ + 3 = 4

⇒ 4 sin² θ = 4 - 3

⇒ 4 sin² θ = 1

⇒ sin² θ = $\frac{1}{4}$

Taking square root of both the sides we get,

⇒ sin θ = $\sqrt{\frac{1}{4}}$

⇒ sin θ = $\frac{1}{2}$

∴ θ = 30°.

Hence, the value of θ = 30°.

mn = (sec θ + tan θ)(sec θ - tan θ)

mn = (sec² θ - tan² θ)

By, trigonometric identities sec² θ - tan² θ = 1.

∴ mn = 1.

Hence, proved that mn = 1.

Given,

x = a sec θ + b tan θ and y = a tan θ + b sec θ.

∴ x² - y² = (a sec θ + b tan θ)² - (tan θ + b sec θ)²

⇒ x² - y² = a² sec² θ + b² tan² θ + 2ab sec θ tan θ - (a² tan² θ + b² sec ² θ + 2ab sec θ tan θ)

⇒ x² - y² = a² sec² θ + b² tan² θ + 2ab sec θ tan θ - a² tan² θ - b² sec ² θ - 2ab sec θ tan θ

⇒ x² - y² = a²(sec² θ - tan² θ) - b²(sec² θ - tan² θ)

⇒ x² - y² = a² - b².

Hence, proved that x² - y² = a² - b².

Given, x = h + a cos θ and y = k + a sin θ

∴ (x - h)² + (y - k)² = (h + a cos θ - h)² + (k + a sin θ - k)²
⇒ (x - h)² + (y - k)² = (a cos θ)² + (a sin θ)²
⇒ (x - h)² + (y - k)² = a² cos² θ + a² sin² θ
⇒ (x - h)² + (y - k)² = a² (cos² θ + sin² θ)
⇒ (x - h)² + (y - k)² = a².

Hence, proved that (x - h)² + (y - k)² = a².

Given,

$\text{cot}^2 \text{ θ} - \frac{1}{\text{sin}^2 \text{ θ}}$

The equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}} - \frac{1}{\text{sin}^2 \text{ θ}} \\[1em] \Rightarrow \frac{\text{cos}^2 \text{ θ} - 1}{\text{sin}^2 \text{ θ}} \\[1em] \Rightarrow \frac{-(1 - \text{cos}^2 \text{ θ})}{\text{sin}^2 \text{ θ}} \\[1em] \Rightarrow \frac{-\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ}} \\[1em] \Rightarrow -1.\end{aligned}$$

Hence, Option 2 is the correct option.

Given, (sec² θ - 1)(1 - cosec² θ)

By using trigonometric identities the equation can be written as,

$$\begin{aligned}\Rightarrow \text{tan}^2 \text{ θ} (-\text{cot}^2 \text{ θ}) \\[1em] \Rightarrow \text{tan}^2 \text{ θ} \times \frac{-1}{\text{tan}^2 \text{ θ}} \\[1em] \Rightarrow -1.\end{aligned}$$

Hence, Option 1 is the correct option.

Given, $\frac{\text{tan}^2 \text{ θ}}{\text{1 + tan}^2 \text{ θ}}$

On solving,

$$\begin{aligned}\Rightarrow \frac{\frac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{1 + \frac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} \\[1em] \Rightarrow \frac{\frac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\frac{\text{cos}^2 \text{ θ} + \text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} \\[1em] \Rightarrow \frac{\text{sin}^2 \text{ θ} \text{ cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ} (\text{cos}^2 \text{ θ} + \text{sin}^2 \text{ θ})} \\[1em] \Rightarrow \text{sin}^2 \text{ θ}.\end{aligned}$$

Hence, Option 3 is the correct option.

Given, (cos θ + sin θ)² + (cos θ - sin θ)²

On solving,

⇒ cos² θ + sin² θ + 2cos θ sin θ + cos² θ + sin² θ - 2cos θ sin θ
⇒ 2(cos² θ + sin² θ)
⇒ 2.

Hence, Option 4 is the correct option.

Given, (sec A + tan A)(1 - sin A)

On solving,

$$\begin{aligned}\Rightarrow \Big(\frac{1}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}}\Big)(1 - \text{sin A}) \\[1em] \Rightarrow \frac{\text{(1 + sin A)(1 - sin A)}}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{1 - sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\text{cos A}} \\[1em] \Rightarrow \text{cos A}.\end{aligned}$$

Hence, Option 4 is the correct option.

Given, $\frac{1 + \text{tan}^2 A}{1 + \text{cot}^2 A}$.

By using trigonometric identities the above equation can be written as,

$$\begin{aligned}\Rightarrow \frac{\text{sec}^2 A}{\text{cosec}^2 A} \\[1em] \Rightarrow \frac{\frac{1}{\text{cos}^2 A}}{\frac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \text{tan}^2 A\end{aligned}$$

Hence, Option 4 is the correct option.

We know that,

⇒ sec² θ - tan² θ = 1

∴ (sec θ - tan θ)(sec θ + tan θ) = 1
⇒ k (sec θ + tan θ) = 1
⇒ (sec θ + tan θ) = $\frac{1}{k}$

Hence, Option 4 is the correct option.

Since, θ is an acute angle triangle,

cos(90° - θ) = sin θ and sin(90° - θ) = cos θ.

Using above values in sin θ cos(90° - θ) + cos θ sin (90° - θ) we get,

⇒ sin θ sin θ + cos θ cos θ
⇒ sin² θ + cos² θ
⇒ 1.

Hence, Option 3 is the correct option.

Since, angles are acute in the equation,

∴ cos(90° - θ) = sin θ and sin(90° - θ) = cos θ

Using above values in cos 65° sin 25° + sin 65° cos 25° we get,

⇒ cos 65° sin (90 - 65)° + sin 65° cos (90 - 65)°
⇒ cos 65° cos 65° + sin 65° sin 65°
⇒ cos² 65° + sin² 65°
⇒ 1.

Hence, Option 2 is the correct option.

Solving 3 tan² 26° - 3 cosec² 64°,

⇒ 3 tan² 26° - 3 cosec² (90 - 26)°
⇒ 3 tan² 26° - 3 sec² 26°
⇒ 3(tan² 26° - sec² 26°)
⇒ 3 × -1
⇒ -3.

Hence, Option 3 is the correct option.

Trigonometry identity :

sin² θ + cos² θ = 1

cosec² θ - cot² θ = 1

∴ Only statement (i) is correct.

Hence, Option 1 is the correct option.

According to assertion, sec² 23° - tan² 23° = 1

Solving L.H.S of above equation,

$$\begin{aligned}\Rightarrow \text{sec}^2 23° - \text{tan}^2 23° \\[1em] \Rightarrow \frac{1}{\text{cos}^2 23°} - \frac{\text{sin}^2 23°}{\text{cos}^2 23°} \\[1em] \Rightarrow \frac{1 - \text{sin}^2 23°}{\text{cos}^2 23°} \\[1em] \Rightarrow \frac{\text{cos}^2 23°}{\text{cos}^2 23°} \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

∴ Assertion (A) is true.

According to standard trigonometric values, the correct evaluation is:

cos 60° = $\frac{1}{2}$
​ ∴ Reason (R) is false.

∴ Assertion (A) is true, Reason (R) is false.

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \text{cosec }^2 θ - \text{cot }^2 θ \\[1em] \Rightarrow \frac{1}{\text{sin }^2 θ} - \frac{\text{cos }^2 θ}{\text{sin }^2 θ} \\[1em] \Rightarrow \frac{1 - \text{cos }^2 θ}{\text{sin }^2 θ}\\[1em] \Rightarrow \frac{\text{sin }^2 θ}{\text{sin }^2 θ}\\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

So, cosec² θ - cot² θ = 1, the condition 0 < θ ≤ 90° ensures that both cosec θ and cot θ are defined and non-zero, making the statement valid.

∴ Reason (R) is true.

⇒ cosec² θ - cot² θ = 1

⇒ (cosec θ - cot θ)(cosec θ + cot θ) = 1

⇒ (cosec θ - cot θ) = $\frac{1}{\text{cosec θ + cot θ}}$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \text{cosec }^2 θ - \text{cot }^2 θ\\[1em] \Rightarrow \frac{1}{\text{sin }^2 θ} - \frac{\text{cos }^2 θ}{\text{sin }^2 θ}\\[1em] \Rightarrow \frac{1 - \text{cos }^2 θ}{\text{sin }^2 θ}\\[1em] \Rightarrow \frac{\text{sin }^2 θ}{\text{sin }^2 θ}\\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

So, cosec² θ - cot² θ = 1, the condition 0 < θ ≤ 90° ensures that both cosec θ and cot θ are defined and non-zero, making the statement valid.

∴ Reason (R) is true.

When θ = 54° and 0° < 54° ≤ 90°

So, cosec² 54° - cot² 54° = 1.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

An equation involving trigonometric ratios is a trigonometric identity if it holds true for all possible values of the angles involved.

∴ Reason (R) is true.

1 + tan² θ = sec² θ is a correct trigonometric identity.

∴ Assertion (A) is false.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

sin θ = $\frac{1}{\text{cosec θ}} = \frac{1}{\sqrt{5}}$,

cos² θ = 1 - sin² θ = 1 - $\Big(\frac{1}{\sqrt{5}}\Big)^2 = 1 - \frac{1}{5} = \frac{4}{5}.$

cos θ = $\sqrt{\frac{4}{5}}$ = $\frac{2}{\sqrt{5}}$.

cot θ = $\frac{\text{cos θ}}{\text{sin θ}} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}$ = 2.

$$\begin{aligned}\text{cot θ - cos θ} = 2 - \frac{2}{\sqrt{5}} \\[1em] = 2\Big(1 - \frac{1}{\sqrt{5}}\Big) \\[1em] = \frac{2(\sqrt{5} - 1)}{\sqrt{5}}.\end{aligned}$$

Hence, the value of cot θ - cos θ = $\frac{2(\sqrt{5} - 1)}{\sqrt{5}}$.

sec² θ = 1 + tan² θ

sec² θ = 1 + $\Big(\frac{8}{15}\Big)^2$

sec² θ = 1 + $\frac{64}{225} = \frac{225 + 64}{225} = \frac{289}{225}$.

sec θ = $\sqrt{\frac{289}{225}} = \frac{17}{15}$.

cot θ = $\frac{1}{\text{tan θ}} = \frac{1}{\frac{8}{15}} = \frac{15}{8}.$

cosec² θ = 1 + cot² θ

cosec² θ = 1 + $\Big(\frac{15}{8}\Big)^2$

cosec² θ = 1 + $\frac{225}{64} = \frac{64 + 225}{64} = \frac{289}{64}$.

cosec θ = $\sqrt{\frac{289}{64}} = \frac{17}{8}$.

$$\begin{aligned}\text{sec θ + cosec θ} = \frac{17}{15} + \frac{17}{8} \\[1em] = \frac{17 \times 8 + 17 \times 15}{120} \\[1em] = \frac{136 + 255}{120} \\[1em] = \frac{391}{120} \\[1em] = 3\frac{31}{120}.\end{aligned}$$

Hence, the value of expression sec θ + cosec θ = $3\frac{31}{120}.$

Since, angles are acute in the equation,

∴ cos(90° - θ) = sin θ, sin(90° - θ) = cos θ and tan(90° - θ) = cot θ.

Using above equations in

2 x $\Big(\frac{\text{cos}^2 20° + \text{cos}^2 70°}{\text{sin}^2 25° + \text{sin}^2 65°}\Big)$ - tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

= 2 x $\Big(\frac{\text{cos}^2 20° + \text{cos}^2 (90 - 20)°}{\text{sin}^2 25° + \text{sin}^2 (90 - 25)°}\Big)$ - tan 45° + tan 13° tan 23° tan 30° tan (90 - 23)° tan (90 - 13)°

= 2 x $\Big(\frac{\text{cos}^2 20° + \text{sin}^2 20°}{\text{sin}^2 25° + \text{cos}^2 25°}\Big)$ - tan 45° + tan 13° tan 23° tan 30° cot 23° cot 13°

= 2 x 1 - 1 + tan 13° cot 13° tan 23° cot 23° tan 30°

= 1 + 1 x $\frac{1}{\sqrt{3}}$

= $\frac{\sqrt{3} + 1}{\sqrt{3}}$

= $\frac{(\sqrt{3} + 1)\sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

= $\frac{3 + \sqrt{3}}{\sqrt{3}}$.

Hence, the value of the expression is $\frac{3 + \sqrt{3}}{\sqrt{3}}$.

Since, angles are acute in the equation,

∴ cos(90° - θ) = sin θ, sin(90° - θ) = cos θ.

Using above equations in

$\frac{\text{sin}^2 22° + \text{sin}^2 68°}{\text{cos}^2 22° + \text{cos}^2 68°}$ + sin² 63° + cos 63° sin 27°

= $\frac{\text{sin}^2 22° + \text{sin}^2 (90 - 22)°}{\text{cos}^2 22° + \text{cos}^2 (90 - 22)°}$ + sin² 63° +cos 63° sin (90 - 63)°

= $\frac{\text{sin}^2 22° + \text{cos}^2 22°}{\text{cos}^2 22° + \text{sin}^2 22°}$ + sin² 63° + cos 63° cos 63°

= $\frac{1}{1}$ + sin² 63° + cos² 63°

= 1 + 1

= 2.

Hence, the value of expression is 2.

Solving the L.H.S of above equation using trigonometric identities,

⇒ $\frac{4}{3}$[sec² 59° - cot² (90 - 59)°] - $\frac{2}{3}$ sin 90° + 3 tan² 56° tan² (90 - 56)°

= $\frac{4}{3}$(sec² 59° - tan² 59°) - $\frac{2}{3}$ sin 90° + 3 tan² 56° cot² 56°

= $\frac{4}{3}$ x 1 - $\frac{2}{3}$ x 1 + 3 x 1

= $\frac{4}{3}$ - $\frac{2}{3}$ + 3

= $\frac{2}{3}$ + 3

= $\frac{11}{3}$.

Comparing it with R.H.S i.e.,

$\frac{11}{3} = \frac{x}{3}$

x = 11.

Hence, the value of x = 11.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{\text{cos A(1 + sin A) + \text{cos A(1 - sin A)}}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] = \frac{\text{cos A + cos A sin A + cos A - cos A sin A}}{1 - \text{sin}^2 A} \\[1em] = \frac{2\text{ cos A}}{\text{cos}^2 A} \\[1em] = \frac{2}{\text{cos A}} \\[1em] = 2\text{ sec A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{cos A}}{\text{1 - sin A}} + \frac{\text{cos A}}{\text{1 + sin A}} = 2\text{ sec A}$.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{\text{cos A(cosec A - 1) + \text{cos A(cosec A + 1)}}}{\text{(cosec A - 1)(cosec A + 1)}} \\[1em] = \frac{\text{cos A cosec A - cos A + cos A cosec A + cos A}}{\text{cosec}^2 A - 1} \\[1em] = \frac{2\text{ cos A } \times \frac{1}{\text{sin A}}}{\text{cot}^2 A} \\[1em] = \frac{2\text{cot A}}{\text{cot}^2 A} \\[1em] = \frac{2}{\text{cot A}} \\[1em] = 2 \text{ tan A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{cos A}}{\text{cosec A + 1}} + \frac{\text{cos A}}{\text{cosec A - 1}} = 2 \text{ tan A}$.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{\text{(cos θ - sin θ)}(1 + \frac{\text{sin θ}}{\text{cos θ}})}{\text{2 cos}^2 \text{ θ} - 1} \\[1em] = \frac{\frac{\text{(cos θ - sin θ)(cos θ + sin θ)}}{\text{cos θ}}}{\text{2 cos}^2 \text{ θ} - 1} \\[1em] = \frac{\text{cos}^2 \text{ θ} - \text{sin}^2 θ}{\text{cos θ}(\text{2 cos}^2 \text{ θ} - 1)} \\[1em] = \frac{\text{cos}^2 \text{ θ} - (1 - \text{cos}^2 \text{ θ})}{\text{cos θ(2 cos}^2 \text{ θ} - 1)} \\[1em] = \frac{\text{cos}^2 \text{ θ} + \text{cos}^2 \text{ θ} - 1}{\text{cos θ(2 cos}^2 \text{ θ} - 1)} \\[1em] = \frac{\text{(2 cos}^2 \text{ θ} - 1)}{\text{cos θ(2 cos}^2 \text{ θ} - 1)} \\[1em] = \frac{1}{\text{cos θ}} \\[1em] = \text{sec θ}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that $\frac{\text{(cos θ - sin θ)(1 + tan θ)}}{2\text{ cos}^2 \text{ θ} - 1} = \text{sec θ}$.

Solving L.H.S.,

⇒ sin² θ + cos⁴ θ

= 1 - cos² θ + (cos² θ)²

= 1 - cos² θ + (1 - sin² θ)²

= 1 - cos² θ + 1 + sin⁴ θ - 2sin² θ

= 1 - cos² θ + 1 + sin⁴ θ - 2(1 - cos² θ)

= 2 - cos² θ + sin⁴ θ - 2 + 2cos² θ

= 2cos² θ - cos² θ + sin⁴ θ

= cos² θ + sin⁴ θ.

Since, L.H.S. = R.H.S. hence, proved that sin² θ + cos⁴ θ = cos² θ + sin⁴ θ.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{\text{cot}^2 \text{ θ} + \text{(cosec θ + 1)}^2}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \frac{\text{cot}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + 1 + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \frac{\text{cot}^2 \text{ θ} + 1 + \text{cosec}^2 \text{ θ} + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \frac{\text{cosec}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \frac{2\text{ cosec}^2 \text{ θ} + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \frac{\text{2 cosec θ(cosec θ + 1)}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \frac{\text{2 cosec θ}}{\text{cot θ}} \\[1em] = \frac{\frac{2}{\text{sin θ}}}{\frac{\text{cos θ}}{\text{sin θ}}} \\[1em] = \frac{2}{\text{cos θ}} \\[1em] = 2\text{ sec θ}.\end{aligned}$$

Since, L.H.S. = R.H.S hence proved that $\frac{\text{cot θ}}{\text{cosec θ + 1}} + \frac{\text{cosec θ + 1}}{\text{cot θ}} = 2\text{ sec θ}$.

Solving,

$$\begin{aligned}\phantom{\Rightarrow} \text{(sin θ + cos θ)(cosec θ - sec θ)} \\[1em] \Rightarrow \text{(sin θ + cos θ)} \times \Big(\frac{1}{\text{sin θ}} - \frac{1}{\text{cos θ}}\Big) \\[1em] \Rightarrow \text{(sin θ + cos θ)} \times \Big(\frac{\text{cos θ - sin θ}}{\text{sin θ cos θ}}\Big) \\[1em] \Rightarrow \frac{\text{cos}^2 θ - \text{sin}^2 θ}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \frac{\text{1 - 2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \quad [\because \text{cos}^2 \text{ θ} = 1 - \text{sin}^2 \text{ θ}] \\[1em] \Rightarrow \frac{1}{\text{sin θ.cos θ}} - \frac{\text{2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow \text{cosec θ.sec θ} - \frac{\text{2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow \text{cosec θ.sec θ - 2 tan θ}.\end{aligned}$$

Hence, proved that (sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{1}{\text{cos}^4 A}\text{(1 + sin }^2 A)\text{(1 - sin}^2 A) - \frac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \frac{\text{(1 + sin}^2 A)\text{cos}^2 A}{{\text{cos}^4 A}} - \frac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \frac{1 + \text{sin}^2 A}{\text{cos}^2 A} - \frac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \frac{1 + \text{sin}^2 A - 2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \frac{1 - \text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \frac{\text{cos}^2 A}{\text{cos}^2 A} \\[1em] = 1.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that sec⁴ A(1 - sin⁴ A) - 2tan² A = 1.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin A + cos A + 1}} + \frac{1}{\text{sin A + cos A - 1}} \\[1em] = \frac{\text{sin A + cos A - 1 + sin A + cos A + 1}}{\text{(sin A + cos A + 1)}\text{(sin A + cos A - 1)}} \\[1em] = \frac{2\text{(sin A + cos A)}}{\text{(sin A + cos A)}^2 - 1} \\[1em] = \frac{2\text{(sin A + cos A)}}{\text{sin}^2 A + \text{cos}^2 A + \text{2 sin A cos A} - 1} \\[1em] = \frac{\text{2 sin A + 2 cos A}}{1 - 1 + \text{2 sin A cos A}} \\[1em] = \frac{\text{2 sin A + 2 cos A}}{\text{2 sin A cos A}} \\[1em] = \frac{\text{2 sin A}}{\text{2 sin A cos A}} + \frac{\text{2 cos A}}{\text{2 sin A cos A}} \\[1em] = \frac{1}{\text{cos A}} + \frac{1}{\text{sin A}} \\[1em] = \text{sec A + cosec A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that $\frac{1}{\text{sin A + cos A + 1}} + \frac{1}{\text{sin A + cos A - 1}} = \text{sec A + cosec A}$.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} \\[1em] = \frac{\text{(sin θ + cos θ)}(\text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} - \text{sin θ cos θ})}{\text{sin θ + cos θ}} + \text{sin θ cos θ} \\[1em] = \text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} - \text{sin θ cos θ + sin θ cos θ} \\[1em] = 1.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that $\frac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1$.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \Big(\frac{1}{\text{cos A}} - \frac{\text{sin A}}{\text{cos A}}\Big)^2(1 + \text{sin A}) \\[1em] = \Big(\frac{\text{1 - sin A}}{\text{cos A}}\Big)^2\text{(1 + sin A)} \\[1em] = \frac{\text{(1 - sin A)}^2\text{(1 + sin A)}}{\text{1 - sin}^2 A} \\[1em] = \frac{\text{(1 - sin A)}^2\text{(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] = \text{1 - sin A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that (sec A - tan A)²(1 + sin A) = 1 - sin A.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{\text{cos A}}{1 - \frac{\text{sin A}}{\text{cos A}}} - \frac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \frac{\text{cos A}}{\frac{\text{cos A - sin A}}{\text{cos A}}} - \frac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \frac{\text{cos}^2 A}{\text{cos A - sin A}} - \frac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \frac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \frac{\text{(cos A - sin A)(cos A + sin A)}}{\text{(cos A - sin A)}} \\[1em] = \text{cos A + sin A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that $\frac{\text{cos A}}{\text{1 - tan A}} - \frac{\text{sin}^2 A}{\text{cos A - sin A}} = \text{sin A + cos A}$.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \text{(sec A - cosec A)(1 + tan A + cot A)} \\[1em] = \Big(\frac{1}{\text{cos A}} - \frac{1}{\text{sin A}}\Big)\Big(1 + \frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}}\Big) \\[1em] = \Big(\frac{\text{sin A - cos A}}{\text{sin A cos A}}\Big)\Big(\frac{\text{sin A cos A} + \text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}} \Big) \\[1em] = \frac{(\text{sin A - cos A})(\text{sin A cos A + 1})}{\text{sin}^2 A \space \text{cos}^2 A} \\[1em]\end{aligned}$$

Now solving R.H.S.,

$$\begin{aligned}\Rightarrow \text{tan A sec A - cot A cosec A} \\[1em] = \frac{\text{sin A}}{\text{cos A}} \times \frac{1}{\text{cos A}} - \frac{\text{cos A}}{\text{sin A}} \times \frac{1}{\text{sin A}} \\[1em] = \frac{\text{sin A}}{\text{cos}^2 A} - \frac{\text{cos A}}{\text{sin}^2 A} \\[1em] = \frac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin}^2 A \text{ cos}^2 A} \\[1em] = \frac{\text{(sin A - cos A)(sin}^2 A + \text{cos}^2 A + \text{sin A cos A})}{\text{sin}^2 A \text{ cos}^2 A} \\[1em] = \frac{(\text{sin A - cos A})(\text{sin A cos A + 1})}{\text{sin}^2 A \space \text{cos}^2 A}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that (sec A - cosec A)(1 + tan A + cot A) = tan A sec A - cot A cosec A.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{\frac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\frac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}} - 1} + \frac{\frac{1}{\text{sin}^2 \text{ θ}}}{\frac{1}{\text{cos}^2 \text{ θ}} - \frac{1}{\text{sin}^2 \text{ θ}}} \\[1em] = \frac{\frac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\frac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} + \frac{\frac{1}{\text{sin}^2 \text{ θ}}}{\frac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ} \text{ sin}^2 \text{ θ}}} \\[1em] = \frac{\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} + \frac{1}{\frac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} \\[1em] = \frac{\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} + \frac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} \\[1em] = \frac{\text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} \\[1em] = \frac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

$\frac{\text{tan}^2 \text{ θ}}{\text{tan}^2 \text{ θ} - 1} + \frac{\text{cosec}^2 \text{ θ}}{\text{sec}^2 \text{ θ} - \text{cosec}^2 \text{ θ}} = \frac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}$.

Given,

$$\begin{aligned}\Rightarrow \frac{\text{sin A + cos A}}{\text{sin A - cos A}} + \frac{\text{sin A - cos A}}{\text{sin A + cos A}}\\[1em] \Rightarrow \frac{\text{(sin A + cos A)}^2 + \text{(sin A - cos A)}^2}{(\text{sin} A - \text{cos} A)(\text{sin A + cos A})}\\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A + 2\text{sin A.cos A} + \text{sin}^2 A + \text{cos}^2 A - 2\text{sin A.cos A}}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \frac{2\text{sin}^2 A + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \frac{2(1 - \text{cos}^2 A) + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \frac{2 - 2\text{cos}^2 A + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \frac{2}{\text{sin}^2 A - \text{cos}^2 A}\end{aligned}$$

So proved,

$\frac{\text{sin A + cos A}}{\text{sin A - cos A}} + \frac{\text{sin A - cos A}}{\text{sin A + cos A}} = \frac{2}{\text{sin }^2A - \text{cos }^2A}$

Now,

$$\begin{aligned}\Rightarrow \frac{2}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \frac{2}{(1 - \text{cos}^2 A) - \text{cos}^2 A}\\[1em] \Rightarrow \frac{2}{1 - \text{cos}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \frac{2}{1 - 2\text{cos}^2 A}\end{aligned}$$

So proved,

$\frac{\text{sin A + cos A}}{\text{sin A - cos A}} + \frac{\text{sin A - cos A}}{\text{sin A + cos A}} = \frac{2}{1 - 2\text{cos}^2 A}$

Now,

$\Rightarrow \frac{2}{\text{sin}^2 A - \text{cos}^2 A}$

Dividing numerator and denominator by cos² A, we get :

$$\begin{aligned}\Rightarrow \frac{\frac{2}{\text{cos}^2 A}}{\frac{\text{sin}^2 A}{\text{cos}^2 A} - \frac{\text{cos}^2 A}{\text{cos}^2 A}}\\[1em] \Rightarrow \frac{2\text{sec}^2 A}{\text{tan}^2 A - 1}\\[1em]\end{aligned}$$

So proved,

$\frac{\text{sin A + cos A}}{\text{sin A - cos A}} + \frac{\text{sin A - cos A}}{\text{sin A + cos A}} = \frac{2\text{sec }^2A}{\text{tan }^2A - 1}$

Hence, proved that

$\frac{\text{sin A + cos A}}{\text{sin A - cos A}} + \frac{\text{sin A - cos A}}{\text{sin A + cos A}} = \frac{2}{\text{sin }^2A - \text{cos }^2A} = \frac{2}{1 - 2\text{cos }^2A} = \frac{2\text{sec }^2A}{\text{tan }^2A - 1}$.

Solving L.H.S.,

⇒ 2[(sin² θ)³ + (cos² θ)³] - 3[(sin² θ)² + (cos² θ)²] + 1

⇒ 2[(sin² θ + cos² θ)³ - 3sin² θ cos² θ(sin² θ + cos² θ)] - 3[(sin² θ + cos² θ)² - 2sin² θ cos² θ)] + 1

⇒ 2[(1)³ - 3sin² θ cos² θ(1)] - 3[(1)² - 2sin² θ cos² θ)] + 1

⇒ 2 - 6sin² θ cos² θ - 3 + 6sin² θ cos² θ + 1

⇒ 2 - 3 + 1 - 6sin² θ cos² θ + 6sin² θ cos² θ

⇒ 3 - 3

⇒ 0.

Since, L.H.S. = R.H.S. hence, proved that 2(sin⁶ θ + cos⁶ θ) - 3(sin⁴ θ + cos⁴ θ) + 1 = 0.

Given,

cot θ + cos θ = m ....(i)
cot θ - cos θ = n ....(ii)

Adding (i) and (ii) we get,

⇒ m + n = cot θ + cos θ + cot θ - cos θ
⇒ m + n = 2 cot θ
⇒ 2 cot θ = m + n
⇒ cot θ = $\frac{m + n}{2}$.

∴ tan θ = $\frac{2}{m + n}$ ....(iii)

Subtracting (ii) from (i) we get,

m - n = cot θ + cos θ - cot θ + cos θ
m - n = 2 cos θ
cos θ = $\frac{m - n}{2}$.

∴ sec θ = $\frac{2}{m - n}$ ....(iv)

Squaring and subtracting (iii) from (iv),

$$\begin{aligned}\Rightarrow \text{sec}^2 \text{ θ} - \text{tan}^2 \text{ θ} = \Big(\frac{2}{m - n}\Big)^2 - \Big(\frac{2}{m + n}\Big)^2 \\[1em] \Rightarrow 1 = \frac{4}{(m - n)^2} - \frac{4}{(m + n)^2} \\[1em] \Rightarrow 4\Big[\frac{1}{(m - n)^2} - \frac{1}{(m + n)^2}\Big] = 1 \\[1em] \Rightarrow 4\Big[\frac{(m + n)^2 - (m - n)^2}{(m + n)^2(m - n)^2}\Big] = 1 \\[1em] \Rightarrow 4\Big[\frac{m^2 + n^2 + 2mn - m^2 - n^2 + 2mn}{(m + n)(m - n)(m + n)(m - n)} \Big] = 1 \\[1em] \Rightarrow 4\Big[\frac{\cancel{m^2} + \cancel{n^2} + 2mn - \cancel{m^2} - \cancel{n^2} + 2mn}{(m^2 - n^2)(m^2 - n^2)} \Big] = 1 \\[1em] \Rightarrow 4 \times \frac{4mn}{(m^2 - n^2)^2} = 1 \\[1em] \Rightarrow 16 mn = (m^2 - n^2)^2.\end{aligned}$$

Hence, proved that (m² - n²)² = 16 mn.

Given,

2 cos² θ + sin θ - 2 = 0

On Solving,

⇒ 2(1 - sin² θ) + sin θ - 2 = 0

= 2 - 2 sin² θ + sin θ - 2 = 0

= sin θ -2 sin² θ = 0

= sin θ (1 - 2 sin θ) = 0

So, either sin θ = 0 or 1 - 2 sin θ = 0

If, sin θ = 0
sin θ = sin 0°
θ = 0°.

Given, θ > 0° hence, θ = 0° is not possible.

∴ 1 - 2 sin θ = 0

⇒ 1 = 2 sin θ

⇒ sin θ = $\frac{1}{2}$

⇒ sin θ = sin 30°

⇒ θ = 30°.

Hence, the value of θ = 30°.

Given,

3 cos θ = 2 sin² θ

On Solving,

⇒ 3 cos θ = 2(1 - cos² θ)

⇒ 3 cos θ = 2 - 2cos² θ

⇒ 2 cos² θ + 3 cos θ - 2 = 0

⇒ 2 cos² θ + 4 cos θ - cos θ - 2 = 0

⇒ 2 cos θ(cos θ + 2) - 1(cos θ + 2) = 0

⇒ (2 cos θ - 1)(cos θ + 2) = 0

⇒ 2 cos θ - 1 = 0 or cos θ + 2 = 0

⇒ cos θ = $\frac{1}{2}$ or cos θ = -2.

But cos θ = -2 is not possible.

∴ cos θ = $\frac{1}{2}$

⇒ cos θ = cos 60°

⇒ θ = 60°.

Hence, the value of θ = 60°.

Given,

sec² θ - 2 tan θ = 0

On Solving,

⇒ 1 + tan² θ - 2 tan θ = 0

⇒ tan² θ - 2 tan θ + 1 = 0

⇒ (tan θ - 1)² = 0

⇒ tan θ - 1 = 0

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, the value of θ = 45°.

Given,

tan² θ = 3 (sec θ - 1)

On Solving,

⇒ sec² θ - 1 = 3 sec θ - 3

⇒ sec² θ - 1 - 3 sec θ + 3 = 0

⇒ sec² θ - 3 sec θ + 2 = 0

⇒ sec² θ - 2 sec θ - sec θ + 2 = 0

⇒ sec θ (sec θ - 2) - 1(sec θ - 2) = 0

⇒ (sec θ - 1)(sec θ - 2) = 0

⇒ sec θ - 1 = 0 or sec θ - 2 = 0

⇒ sec θ = 1 or sec θ = 2.

If, sec θ = 1
sec θ = sec 0°
θ = 0°.

Given, θ > 0° hence, θ = 0° is not possible.

∴ sec θ = 2

⇒ sec θ = sec 60°

⇒ θ = 60°.

Hence, the value of θ = 60°.