Trigonometrical Ratios

(i) sin x = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{q}{r}$

(ii) tan x = $\frac{\text{perpendicular}}{\text{base}} = \frac{q}{p}$

(iii) sec x = $\frac{\text{hypotenuse}}{\text{base}} = \frac{r}{p}$

(iv) cos y = $\frac{\text{base}}{\text{hypotenuse}} = \frac{b}{n}$

(v) cot y = $\frac{\text{base}}{\text{perpendicular}} = \frac{b}{m}$

(vi) cosec y = $\frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{n}{m}$

(vii) sin z = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{u}{n}$

(viii) cos z = $\frac{\text{base}}{\text{hypotenuse}} = \frac{k}{n}$

(ix) tan z = $\frac{\text{perpendicular}}{\text{base}} = \frac{u}{k}$

In triangle ABC,

By pythagoras theorem,

AC² = AB² + BC²

AC² = 4² + 3²

AC² = 16 + 9

AC² = 25

AC = $\sqrt{25}$

AC = 5 units

(i) sin A = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{3}{5}$

(ii) cos A = $\frac{\text{base}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{4}{5}$

(iii) cot A = $\frac{\text{base}}{\text{perpendicular}} = \frac{AB}{BC} = \frac{4}{3}$

(iv) sin C = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{4}{5}$

(v) sec C = $\frac{\text{hypotenuse}}{\text{base}} = \frac{AC}{BC} = \frac{5}{3}$

(vi) tan C = $\frac{\text{perpendicular}}{\text{base}} = \frac{AB}{BC} = \frac{4}{3}$

Given a right triangle ABC with hypotenuse BC = 17 units and AB = 15 units.

First, find AC using the Pythagoras theorem :

BC² = AB² + AC²

AC² = BC² - AB²

AC² = 17² - 15²

AC² = 289 - 225

AC² = 64

AC = $\sqrt{64}$

AC = 8 units

(i) sin B = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{AC}{BC} = \frac{8}{17}$

(ii) tan B = $\frac{\text{perpendicular}}{\text{base}} = \frac{AC}{AB} = \frac{8}{15}$

(iii) cos C = $\frac{\text{base}}{\text{hypotenuse}} = \frac{AC}{BC} = \frac{8}{17}$

(iv) cot C = $\frac{\text{base}}{\text{perpendicular}} = \frac{AC}{AB} = \frac{8}{15}$

(v) We have to find

sin B cos C + cos B sin C

First we will find the values of cos B & sin C

cos B = $\frac{\text{base}}{\text{hypotenuse}} = \frac{AB}{BC} = \frac{15}{17}$

sin C = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{AB}{BC} = \frac{15}{17}$

Substituting the values, we get :

$$\begin{aligned}\text{sin B cos C + cos B sin C} = \frac{8}{17}\times \frac{8}{17} + \frac{15}{17} \times \frac{15}{17} \\[1em] = \frac{64}{289} + \frac{225}{289} \\[1em] = \frac{289}{289} \\[1em] = 1.\end{aligned}$$

Hence, sin B cos C + cos B sin C = 1.

(vi) We have to find out

sec²C - tan²C

First we will find out the values of sec C & tan C

sec C = $\frac{\text{hypotenuse}}{\text{base}} = \frac{BC}{AC} = \frac{17}{8}$

tan C = $\frac{\text{perpendicular}}{\text{base}} = \frac{AB}{AC} = \frac{15}{8}$

Now putting the values of sec C & tan C

sec²C - tan²C

= $\Big(\frac{17}{8}\Big)^2 - \Big(\frac{15}{8}\Big)^2$

= $\frac{289}{64} - \frac{225}{64} = \frac{289 - 225}{64}$

= $\frac{64}{64}$

= 1.

Hence, sec²C - tan²C = 1.

Using pythagoras theorem in right angled triangle ADB

AB² = BD² + AD²

AD² = AB² - BD²

AD² = 13² - 5²

AD² = 169 - 25

AD² = 144

AD = $\sqrt{144}$

AD = 12 cm

Now we will find out AC using pythagoras theorem in right angled triangle ADC,

AC² = DC² + AD²

AC² = 16² + 12²

AC² = 256 + 144

AC² = 400

AC = $\sqrt{400}$

AC = 20 cm

Now,

(i) sin B = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{AD}{AB} = \frac{12}{13}$

(ii) sec B = $\frac{\text{hypotenuse}}{\text{base}} = \frac{AB}{BD} = \frac{13}{5}$

(iii) cot B = $\frac{\text{base}}{\text{perpendicular}} = \frac{BD}{AD} = \frac{5}{12}$

(iv) cos C = $\frac{\text{base}}{\text{hypotenuse}} = \frac{DC}{AC} = \frac{16}{20} = \frac{4}{5}$

(v) cosec C = $\frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{AC}{AD} = \frac{20}{12} = \frac{5}{3}$

(vi) tan C = $\frac{\text{perpendicular}}{\text{base}} = \frac{AD}{DC} = \frac{12}{16} = \frac{3}{4}$

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}}$

Let perpendicular = x and hypotenuse = $\sqrt{2}x$

By using Pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = $(\sqrt{2}x)^2$ - x²

Base² = 2x² - x²

Base² = x²

Base = x

Now, calculating the remaining trigonometric ratios :

cos θ = $\frac{\text{base}}{\text{hypotenuse}} = \frac{x}{\sqrt{2}x} = \frac{1}{\sqrt{2}}$

tan θ = $\frac{\text{perpendicular}}{\text{base}} = \frac{x}{x} = 1$

cot θ = $\frac{\text{base}}{\text{perpendicular}} = \frac{x}{x} = 1$

sec θ = $\frac{\text{hypotenuse}}{\text{base}} = \frac{\sqrt{2}x}{x} = \sqrt{2}$

cosec θ = $\frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{\sqrt{2}x}{x} = \sqrt{2}$

tan θ = $\frac{\text{Perpendicular}}{\text{Base}} = \frac{8}{15}$

Let perpendicular = 8x and base = 15x

By pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = (15x)² + (8x)²

Hypotenuse² = 225x² + 64x²

Hypotenuse² = 289x²

Hypotenuse = $\sqrt{289x^2}$

Hypotenuse = 17x

Now, calculating the remaining trigonometric ratios :

sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{8x}{17x} = \frac{8}{17}$.

cos θ = $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{15x}{17x} = \frac{15}{17}$

cot θ = $\frac{\text{Base}}{\text{Perpendicular}} = \frac{15x}{8x} = \frac{15}{8}$

sec θ = $\frac{\text{Hypotenuse}}{\text{Base}} = \frac{17x}{15x}= \frac{17}{15}$

cosec θ = $\frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{17x}{8x} = \frac{17}{8}$

cosec θ = $\frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{\sqrt{10}}{1}$.

Let hypotenuse = $\sqrt{10}x$ and perpendicular = x.

By pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = $(\sqrt{10}x)^2$ - x²

Base² = 10x² - x²

Base² = 9x²

Base = $(\sqrt{9x^2})$

Base = 3x

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{x}{\sqrt{10}x} = \frac{1}{\sqrt{10}}$

cos θ = $\frac{\text{base}}{\text{hypotenuse}} = \frac{3x}{\sqrt{10}x} = \frac{3}{\sqrt{10}}$

tan θ = $\frac{\text{perpendicular}}{\text{base}} = \frac{x}{3x} = \frac{1}{3}$

cot θ = $\frac{\text{base}}{\text{perpendicular}} = \frac{3x}{x} = 3$

sec θ = $\frac{\text{hypotenuse}}{\text{base}} = \frac{\sqrt{10}x}{3x} = \frac{\sqrt{10}}{3}$

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{3}{5}$

Let perpendicular = 3x and hypotenuse = 5x

By using Pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = (5x)² - (3x)²

Base² = 25x² - 9x²

Base² = 16x²

Base = 4x

cos θ = $\frac{\text{base}}{\text{hypotenuse}} = \frac{4x}{5x} = \frac{4}{5}$

tan θ = $\frac{\text{perpendicular}}{\text{base}} = \frac{3x}{4x} = \frac{3}{4}$

tan θ = $\frac{\text{perpendicular}}{\text{base}} = \frac{5}{12}$

Let perpendicular = 5x and base = 12x

By pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = (12x)² + (5x)²

Hypotenuse² = 144x² + 25x²

Hypotenuse² = 169x²

Hypotenuse = $\sqrt{169x^2}$

Hypotenuse = 13x

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{5x}{13x} = \frac{5}{13}$

cos θ = $\frac{\text{base}}{\text{hypotenuse}} = \frac{12x}{13x} = \frac{12}{13}$

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}} =\frac{\sqrt{3}}{2}$

Let perpendicular = $\sqrt{3}x$ and hypotenuse = 2x

We will find the value of base using pythagoras theorem

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = (2x)² - $(\sqrt{3}x)^2$

Base² = 4x² - 3x²

Base² = x²

Base = x

cosec θ = $\frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{2x}{\sqrt{3}x} = \frac{2}{\sqrt{3}}$

cot θ = $\frac{\text{base}}{\text{perpendicular}} = \frac{x}{\sqrt{3}x} = \frac{1}{\sqrt{3}}$

Substituting above values in cosec θ + cot θ, we get :

cosec θ + cot θ = $\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}$

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

Hence, cosec θ + cot θ = $\sqrt{3}$.

13 sin θ = 5

sin θ = $\frac{5}{13}$

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}} =\frac{\text{5}}{\text{13}}$

Let perpendicular = 5x and hypotenuse = 13x

By pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = (13x)² - (5x)²

Base² = 169x² - 25x²

Base² = 144x²

Base = $\sqrt{144x^2}$

Base = 12x

cos θ = $\frac{\text{base}}{\text{hypotenuse}} = \frac{12x}{13x} = \frac{12}{13}$

tan θ = $\frac{\text{perpendicular}}{\text{base}} = \frac{5x}{12x} = \frac{5}{12}$

Substituting values, we get :

$$\begin{aligned}\Rightarrow \frac{\text{5 sin θ - 2 cos θ}}{\text{tan θ}} = \frac{5 \times \frac{5}{13} - 2 \times \frac{12}{13}}{\frac{5}{12}} \\[1em] = \frac{\frac{25}{13} - \frac{24}{13}}{\frac{5}{12}} \\[1em] = \frac{\frac{1}{13}}{\frac{5}{12}} \\[1em] = \frac{12}{65}.\end{aligned}$$

Hence, $\frac{\text{5 sin θ - 2 cos θ}}{\text{tan θ}} = \frac{12}{65}$.

cot θ = $\frac{\text{base}}{\text{perpendicular}} = \frac{1}{\sqrt{3}}$

Let base = x and perpendicular = $\sqrt{3}$x

By using pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = (x)² + $(\sqrt{3}x)^2$

Hypotenuse² = x² + 3x²

Hypotenuse² = 4x²

Hypotenuse = $\sqrt{4x^2}$

Hypotenuse = 2x

Now

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{\sqrt{3}x}{2x} = \frac{\sqrt{3}}{2}$

cos θ = $\frac{\text{base}}{\text{hypotenuse}} = \frac{x}{2x} = \frac{1}{2}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(\frac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \frac{1 - \Big(\frac{1}{2}\Big)^2}{2 - \Big(\frac{\sqrt{3}}{2}\Big)^2} \\[1em] = \frac{1 - \frac{1}{4}}{2 - \frac{3}{4}} \\[1em] = \frac{\frac{4 - 1}{4}}{\frac{8 - 3}{4}} \\[1em] = \frac{\frac{3}{4}}{\frac{5}{4}} \\[1em] = \frac{3}{5}.\end{aligned}$$

Hence, $\Big(\frac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \frac{3}{5}$.

sec θ = $\frac{\text{hypotenuse}}{\text{base}} =\frac{13}{5}$

Let hypotenuse = 13x and base = 5x

We will find perpendicular by using pythagoras theorem

Hypotenuse² = Base² + Perpendicular²

Perpendicular² = Hypotenuse² - Base²

Perpendicular² = (13x)² - (5x)²

Perpendicular² = 169x² - 25x²

Perpendicular² = 144x²

Perpendicular = $\sqrt{144x^2}$

Perpendicular = 12x

Now

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}}= \frac{12x}{13x} = \frac{12}{13}$

cos θ = $\frac{\text{base}}{\text{hypotenuse}} = \frac{5x}{13x} = \frac{5}{13}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} = \frac{2 \times \frac{12}{13} - 3 \times \frac{5}{13}}{4 \times \frac{12}{13} - 9 \times \frac{5}{13}} \\[1em] = \frac{\frac{24}{13} - \frac{15}{13}}{\frac{48}{13} - \frac{45}{13}} \\[1em] = \frac{\frac{9}{13}}{\frac{3}{13}} \\[1em] = \frac{9}{3} \\[1em] = 3.\end{aligned}$$

Hence, proved that $\frac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} = 3$.

tan θ = $\frac{\text{perpendicular}}{\text{base}}=\frac{4}{3}$

Let Perpendicular = 4x and Base = 3x

We will find hypotenuse by using pythagoras theorem

Hypotenuse² = Perpendicular² + Base²

Hypotenuse² = (4x)² + (3x)²

Hypotenuse² = 16x² + 9x²

Hypotenuse² = 25x²

Hypotenuse = 5x

Now

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}}= \frac{4x}{5x} = \frac{4}{5}$

cos θ = $\frac{\text{base}}{\text{hypotenuse}}= \frac{3x}{5x} = \frac{3}{5}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{\text{3 sin θ + 2 cos θ}}{\text{3 sin θ - 2 cos θ}} \\[1em] = \frac{3\times\frac{4}{5} + 2\times\frac{3}{5}}{3\times\frac{4}{5} -2\times\frac{3}{5}} \\[1em] = \frac{\frac{12}{5} + \frac{6}{5}}{\frac{12}{5} -\frac{6}{5}}\\[1em] = \frac{\frac{12+6}{5}}{\frac{12-6}{5}} \\[1em] = \frac{\frac{18}{5}}{\frac{6}{5}} \\[1em] = \frac{18\times5}{5\times6}\\[1em] = \frac{18}{6} = 3\end{aligned}$$

Hence, proved that $\Big(\frac{3\sin θ + 2\cos θ}{3\sin θ - 2\cos θ}\Big)$ = 3.

cot θ = $\frac{\text{base}}{\text{perpendicular}} = \frac{q}{p}$

Let base = qx and perpendicular = px

We will find hypotenuse by using pythagoras theorem

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = (qx)² + (px)²

Hypotenuse² = (q² + p²)x²

Hypotenuse = $\sqrt{(q^2 + p^2)x^2}$

Hypotenuse = $\sqrt{(q^2 + p^2)}x$

Now,

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}}= \frac{px}{\sqrt{(p^2 + q^2)}x} = \frac{p}{\sqrt{p^2 + q^2}}$

cos θ = $\frac{\text{base}}{\text{hypotenuse}}= \frac{qx}{\sqrt{(p^2 + q^2)}x} = \frac{q}{\sqrt{p^2 + q^2}}$

Substituting values we get :

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

Hence, proved that $\frac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}$ = $\frac{p^2 - q^2}{p^2 + q^2}$.

cot θ = $\frac{\text{base}}{\text{perpendicular}}=\frac{3}{4}$

Let base = 3x and perpendicular = 4x

We will find hypotenuse by using pythagoras theorem

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = (3x)² + (4x)²

Hypotenuse² = 9x² + 16x²

Hypotenuse² = 25x²

Hypotenuse = 5x

Now

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}}= \frac{4x}{5x} = \frac{4}{5}$

cos θ = $\frac{\text{base}}{\text{hypotenuse}}= \frac{3x}{5x} = \frac{3}{5}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{\sin θ - \cos θ}{\sin θ+ \cos θ}\\[1em] = \frac{\frac{4}{5} -\frac{3}{5}}{\frac{4}{5} + \frac{3}{5}}\\[1em] = \frac{\frac{4-3}{5}}{\frac{4+3}{5}}\\[1em] = \frac{\frac{1}{5}}{\frac{7}{5}}\\[1em] = \frac{1}{5}\times\frac{5}{7}\\[1em] = \frac{1}{7}.\end{aligned}$$

Hence, proved that $\frac{\sin θ - \cos θ}{\sin θ+ \cos θ}$ = $\frac{1}{7}$

In right angled triangle DBC,

Perpendicular = BC = 8 cm

Base = DB = 6 cm

Then we will find hypotenuse (CD) by pythagoras theorem,

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = 6² + 8²

Hypotenuse² = 36 + 64

Hypotenuse² = 100

Hypotenuse = 10 cm

In right angled triangle ABC,

Perpendicular = CB = 8 cm

Hypotenuse = AC = 17 cm

Let AD = m

Base (AB) = AD + DB = m + DB

By pythagoras theorem,

Base² = Hypotenuse² - Perpendicular²

(m + 6)² = 17² - 8²

m² + 36 + 12m = 289 - 64

m² + 36 + 12m = 225

m² + 12m + 36 - 225 = 0

m² + 12m - 189 = 0

m² + 21m - 9m - 189 =0

m(m + 21) - 9(m + 21) = 0

(m + 21)(m - 9) = 0

m = -21 or m = 9

Sicne, length can't be negative.

so, m = 9 cm

AB = m + 6 = 9 + 6 = 15 cm

(i) sin x° = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{8}{17}$.

(ii) cos y° = $\frac{\text{base}}{\text{hypotenuse}}= \frac{DB}{DC} = \frac{6}{10} = \frac{3}{5}$.

(iii) 3 tan x° - 2 sin y° + 4 cos y°

tan x° = $\frac{\text{perpendicular}}{\text{base}} = \frac{BC}{AB} = \frac{8}{15}$

sin y° = $\frac{\text{perpendicular}}{\text{hypotenuse}}= \frac{BC}{DC} = \frac{8}{10} = \frac{4}{5}$

Putting values of tan x°, sin y°, cos y° in 3 tan x° - 2 sin y° + 4 cos y°

= $3\times\frac{8}{15} - 2\times\frac{4}{5} + 4\times\frac{3}{5}$

= $\frac{8}{5} - \frac{8}{5} + \frac{12}{5}$

= $\frac{12}{5} = 2\frac{2}{5}$.

In right angled triangle ABC,

Hypotenuse = AC = 13 units

Perpendicular = BC = 5 units

By pythagoras theorem,

Base² = Hypotenuse² - Perpendicular²

Base² = 13² - 5²

Base² = 169 - 25

Base² = 144

Base = 12 units

AB = 12 units.

Draw a perpendicular CE on AD.

In triangle CED,

CE = AB = 12 units and AE = BC = 5 units

From figure,

AD = DE + AE

DE = AD - AE = 14 - 5 = 9 units

In Triangle CED,

By pythagoras theorem,

CD² = CE² + ED²

CD² = 12² + 9²

CD² = 144 + 81

CD² = 225

CD = $\sqrt{225}$ = 15 units.

(i) cos θ = $\frac{\text{base}}{\text{hypotenuse}}= \frac{AB}{AC} = \frac{12}{13}$.

(ii) tan Φ = $\frac{\text{perpendicular}}{\text{base}}= \frac{CE}{DE} = \frac{12}{9} = \frac{4}{3}$.

(iii) cosec Φ = $\frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{CD}{CE} = \frac{15}{12} = \frac{5}{4}$.

As, (tan θ + cot θ) = 5

Squaring both sides, we get :

⇒ (tan θ + cot θ)² = 5²

⇒ tan²θ + cot²θ + 2 tan θ cot θ = 25

⇒ tan²θ + cot²θ + $2\tan \theta \times \frac{1}{\tan \theta}$ = 25

⇒ tan²θ + cot²θ + 2 = 25

⇒ tan²θ + cot²θ = 25 - 2

⇒ tan²θ + cot²θ = 23.

Hence, tan²θ + cot²θ = 23.

Given,

(cos θ + sec θ ) = $\frac{5}{2}$

Squaring both sides,

cos²θ + sec²θ + 2cos θ sec θ = $\frac{25}{4}$

As we know cos θ = $\frac{1}{\sec θ }$

cos²θ + sec²θ + 2 = $\frac{25}{4}$

$\Rightarrow \cos^2θ + \sec^2θ = \frac{25}{4} - 2$

$\Rightarrow \frac{25 - 8}{4} = \frac{17}{4}$

Hence, cos²θ + sec²θ = $\frac{17}{4}$.

In the given figure there are two right angled triangles, △ADC and △BDC.

In △ADC,

∠ACD = 60° and AC = 10 m, CD = x m

cos 60° = $\frac{\text{base}}{\text{hypotenuse}}$

$\frac{1}{2} = \frac{CD}{AC}$

$\frac{1}{2} = \frac{x}{10}$

x = 5 m.

In △BDC,

BC = $5\sqrt{2}$ m

CD = x = 5 m

sin y° = $\frac{\text{perpendicular}}{\text{hypotenuse}}= \frac{DC}{BC} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$

sin y° = $\frac{1}{\sqrt{2}}$

sin y° = sin 45°

y° = 45°.

Hence, x = 5 m and y° = 45°.

tan A = $\frac{\text{perpendicular}}{\text{base}} = \frac{BC}{AB}$

Given,

tan A = $\frac{3}{4}$

Let BC = 3x and AB = 4x.

Now by pythagoras theorem

AC² = BC² + AB²

(20)² = (3x)² + (4x)²

400 = 9x² + 16x²

25x² = 400

x² = 16

x = $\sqrt{16}$ = 4

AB = 4x = 16 cm and BC = 3x = 12 cm.

Hence, length of AB = 16 cm and BC = 12 cm.

cos θ = $\frac{\text{base}}{\text{hypotenuse}} = \frac{2x}{1 + x^2}$

Let base = 2x and hypotenuse = 1 + x²

Now we will find perpendicular by using pythagoras theorem

Perpendicular² = Hypotenuse² - Base²

Perpendicular² = (1 + x²)² - (2x)²

Perpendicular² = 1 + x⁴ + 2x² - 4x²

Perpendicular² = 1 + x⁴ - 2x²

Perpendicular² = (x² - 1)²

Perpendicular = (x² - 1)

Now,

sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{x^2 - 1}{1 + x^2}$

tan θ = $\frac{\text{perpendicular}}{\text {base}} = \frac{x^2 - 1}{2x}$

(i) sin 60° cos 30° + cos 60° sin 30°

= $\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2} + \frac{1}{2}\times\frac{1}{2}$

= $\frac{3}{4} + \frac{1}{4}$

= $\frac{3+1}{4} = \frac{4}{4}$

= 1.

Hence, sin 60° cos 30° + cos 60° sin 30° = 1.

(ii) sin 45° cos 30° - cos 45° sin 30°

= $\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times\frac{1}{2}$

= $\frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}$

= $\frac{\sqrt{3}-1}{2\sqrt{2}}$.

Hence, sin 45° cos 30° - cos 45° sin 30° = $\frac{\sqrt{3}-1}{2\sqrt{2}}$.

(iii) cos 60° cos 45° + sin 60° sin 45°

= $\frac{1}{2}\times\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2}\times\frac{1}{\sqrt2}$

= $\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}$

= $\frac{1 + \sqrt{3}}{2\sqrt{2}}$.

Hence, cos 60° cos 45° + sin 60° sin 45° = $\frac{1 + \sqrt{3}}{2\sqrt{2}}$.

(iv) cos 90° + cos² 45° sin 30° tan 45°

As, cos² 45° = (cos 45°)² = $\Big(\frac{1}{\sqrt{2}}\Big)^2 = \frac{1}{2}$

Therefore,

cos 90° + cos² 45° sin 30° tan 45°

= 0 + $\frac{1}{2}\times\frac{1}{2}\times1$

= $\frac{1}{4}$.

Hence, cos 90° + cos² 45° sin 30° tan 45° = $\frac{1}{4}$.

(i) As,

sin² 45° = (sin 45°)² = $\Big(\frac{1}{\sqrt{2}}\Big)^2 = \frac{1}{2}$

and

cos² 60° = (cos 60°)² = $\Big(\frac{1}{2}\Big)^2 = \frac{1}{4}$

Substituting values we get :

(3sin² 45° + 2cos²60°)

= $3\times\frac{1}{2} + 2\times\frac{1}{4}$

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

= $\frac{3+1}{2} = \frac{4}{2}$ = 2.

Hence, 3 sin² 45° + 2 cos²60° = 2.

(ii) As,

cos² 30° = (cos 30°)² = $\Big(\frac{\sqrt{3}}{2}\Big)^2 = \frac{3}{4}$

and

tan² 60° = (tan 60°)² = $(\sqrt{3})^2$ = 3

Therefore,

3cos² 30° + tan²60°

= $3\times \frac{3}{4}$ + 3

= $\frac{9}{4}$ + 3

= $\frac{9 + 12}{4} = \frac{21}{4} = 5\frac{1}{4}$.

Hence, 3cos² 30° + tan²60° = $5\frac{1}{4}$.

(iii) (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)

$$\begin{aligned}= \Big( 1 + \frac{1}{\sqrt{2}} + \frac{1}{2} \Big)\Big(1 - \frac{1}{\sqrt{2}} + \frac{1}{2}\Big)\\[1em] = \Big(\frac{2\sqrt{2}+ 2 + \sqrt{2}}{2\sqrt{2}}\Big)(\frac{2\sqrt{2} - 2 + \sqrt{2}}{2\sqrt{2}}\Big)\\[1em] = \Big(\frac{3\sqrt{2}+ 2}{2\sqrt{2}}\Big)\Big(\frac{3\sqrt{2}- 2}{2\sqrt{2}}\Big)\\[1em] = \Big(\frac{(3\sqrt{2})^2- (2)^2 }{2\sqrt{2}\times {2\sqrt{2}}}\Big)\\[1em] = \Big(\frac{18-4}{8}\Big) \\[1em] = \frac{14}{8} \\[1em] = \frac{7}{4} \\[1em] = 1\frac{3}{4}.\end{aligned}$$

Hence, (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°) = $1\frac{3}{4}$.

(iv) $2\sqrt{2}$ cos 45°cos 60° + 2 $\sqrt{3}$ sin 30° tan 60° - cos 0°

= 2 $\sqrt{2}\times\frac{1}{\sqrt{2}}\times\frac{1}{2} + 2\sqrt{3}\times\frac{1}{2}\times\sqrt{3} - 1$

= 1 + 3 - 1

= 3.

Hence, 2 $\sqrt{2}$ cos 45°cos 60° + 2 $\sqrt{3}$ sin 30° tan 60° - cos 0° = 3.

(v) As,

cos² 60° = (cos 60°)² = $\Big(\frac{1}{2}\Big)^2 = \frac{1}{4}$

and

tan² 30° = (tan 30°)² = $\Big(\frac{1}{\sqrt{3}}\Big)^2 = \frac{1}{3}$

tan² 45° = (tan 45°)² = 1

tan² 60° = (tan 60°)² = ($\sqrt{3}$)² = 3

sin² 60° = (sin 60°)² = $\Big(\frac{\sqrt{3}}{2}\Big)^2 = \frac{3}{4}$

Therefore,

$\frac{4}{3}$ tan²30°+ sin²60° - 3 cos²60°+ $\frac{3}{4}$ tan²60°- 2 tan²45°

$$\begin{aligned}\frac{4}{3}\times\frac{1}{3}+ \frac{3}{4} - 3 \times\frac{1}{4} + \frac{3}{4}\times 3- 2 \times 1 \\[1em] = \frac{4}{9}+ \frac{3}{4} -\frac{3}{4} + \frac{9}{4}- 2 \\[1em] = \frac{4}{9} + \frac{9}{4} - 2 \\[1em] = \frac{16 + 81 - 72}{36}\\[1em] = \frac{25}{36}.\end{aligned}$$

Hence, $\frac{4}{3}$ tan²30°+ sin²60° - 3 cos²60°+ $\frac{3}{4}$ tan²60°- 2 tan²45° = $\frac{25}{36}$.

(vi) As,

cos² 45° = (cos 45°)² = $\Big(\frac{1}{\sqrt{2}}\Big)^2 = \frac{1}{2}$

sin² 45° = (sin 45°)² = $\Big(\frac{1}{\sqrt{2}}\Big)^2 = \frac{1}{2}$

tan² 60° = (tan 60°)² = ($\sqrt{3}$)² = 3

Therefore,

$\frac{\sin^2 45^\circ + \cos^2 45^\circ}{\tan^2 60^\circ}$

= $\frac{\frac{1}{2}+\frac{1}{2}}{3}$

= $\frac{1}{3}$

Hence, $\frac{\sin^2 45^\circ + \cos^2 45^\circ}{\tan^2 60^\circ} = \frac{1}{3}$.

(i) Solving,

$$\begin{aligned}\frac{\sin 30^\circ - \sin 90^\circ +2\times \cos 0^\circ}{\tan 30^\circ \times \tan 60^\circ}\\[1em] = \frac{\frac{1}{2} - 1 +2\times1}{\frac{1}{\sqrt{3}}\times {\sqrt{3}}}\\[1em] = \frac{\frac{1}{2} - 1 + 2}{\frac{1}{\sqrt{3}}\times {\sqrt{3}}}\\[1em] = \frac{\frac{1}{2} + 1}{1}\\[1em] = \frac{3}{2}.\end{aligned}$$

Hence, the required value = $\frac{3}{2}$.

(ii) As,

sin² 30° = (sin 30°)² = $\Big(\frac{1}{2}\Big)^2 = \frac{1}{4}$

cos² 45° = (cos 45°)² = $\Big(\frac{1}{\sqrt{2}}\Big)^2 = \frac{1}{2}$

tan² 30° = (tan 30°)² = $\Big(\frac{1}{\sqrt{3}}\Big)^2 = \frac{1}{3}$

Therefore,

$$\begin{aligned}\frac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ}\\[1em] =\frac{5\times\frac{1}{4} + \frac{1}{2} - 4\times\frac{1}{3}}{2\times\frac{1}{2} \times\frac{\sqrt{3}}{2}+ 1 }\\[1em] = \frac{\frac{5}{4} + \frac{1}{2} -\frac{4}{3}}{\frac{\sqrt{3}}{2}+1}\\[1em] = \frac{\frac{15 + 6 - 16}{12}}{\frac{\sqrt{3}}{2}+1}\\[1em] = \frac{\frac{5}{12}}{\frac{\sqrt{3} + 2}{2}} \\[1em] = \frac{5}{6(2 + \sqrt{3})}.\end{aligned}$$

Hence, $\frac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ} = \frac{5}{6(2 + \sqrt{3})}$.

(iii) As,

tan² 60° = (tan 60°)² = ($\sqrt{3}$)² = 3

cos² 45° = (cos 45°)² = $\Big(\frac{1}{\sqrt{2}}\Big)^2 = \frac{1}{2}$

sec² 30° = (sec 30°)² = $\Big(\frac{2}{\sqrt{3}}\Big)^2 = \frac{4}{3}$

cos² 90° = (cos 90°)² = 0

cot² 30° = (cot 30°)² = ($\sqrt{3}$)² = 3

Therefore,

$$\begin{aligned}\frac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ}\\[1em] = \frac{3 + 4\times\frac{1}{2} + \frac{4}{3} + 0}{2 + 2 - 3}\\[1em] = \frac{3 + 2 + \frac{4}{3} + 0}{1}\\[1em] = 5+ \frac{4}{3}\\[1em] = \frac{19}{3} \\[1em] = 6\frac{1}{3}.\end{aligned}$$

Hence, $\frac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ} = \frac{19}{3}$

(iv) As,

sin⁴ 30° = (sin 30°)⁴ = $\Big(\frac{1}{2}\Big)^4 = \frac{1}{16}$

cos⁴ 60° = (cos 60°)⁴ = $\Big(\frac{1}{2}\Big)^4 = \frac{1}{16}$

cos² 45° = (cos 45°)² = $\Big(\frac{1}{\sqrt{2}}\Big)^2 = \frac{1}{2}$

sin² 90° = (sin 90°)² = 1

Therefore,

4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°)

= $4\Big(\frac{1}{16} + \frac{1}{16}\Big) - 3 \Big(\frac{1}{2} - 1\Big)$

= $4\Big(\frac{2}{16}\Big) - 3 \Big(\frac{-1}{2}\Big)$

= $\frac{1}{2} + \frac{3}{2}$

= 2.

Hence, 4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°) = 2.

(i) cos 60° cos 30° - sin 60° sin 30°

= $\frac{1}{2}\times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\times\frac{1}{2}$

= 0.

Hence, proved that cos 60° cos 30° - sin 60° sin 30° = 0.

(ii) As,

sin² 30° = (sin 30°)² = $\Big(\frac{1}{2}\Big)^2= \frac{1}{4}$

cos² 30° = (cos 30°)² = $\Big(\frac{\sqrt{3}}{2}\Big)^2 = \frac{3}{4}$

Therefore

Left Hand Side :

cos 60° = $\frac{1}{2}$

Right Hand Side :

(1 - 2 sin²30°)

=1 - 2 $\times \frac{1}{4} = \frac{1}{2}$

(2 cos²30° - 1)

= 2 $\times\frac{3}{4}$ - 1 = $\frac{1}{2}$

Hence proved that cos 60° = (1 - 2 sin²30°) = (2 cos²30° - 1).

(iii) Left Hand Side :

tan 30° = $\frac{1}{\sqrt{3}}$

Right Hand Side

$$\begin{aligned}\Big(\frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big) = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3}\times\frac{1}{\sqrt{3}}}\\[1em] =\frac{\frac{3-1}{\sqrt{3}}}{1 + 1}\\[1em] =\frac{\frac{2}{\sqrt{3}}}{2}\\[1em] = \frac{1}{\sqrt{3}}.\end{aligned}$$

Hence proved that tan 30° = $\Big(\frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)$

(i) Left Hand Side :

sin 60° cos 30° - cos 60° sin 30°

= $\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2} -\frac{1}{2}\times \frac{1}{2}$

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

= $\frac{1}{2}$.

Right Hand Side :

sin 30° = $\frac{1}{2}$

Hence, proved that sin 60° cos 30° - cos 60° sin 30° = sin 30°.

(ii) Left Hand Side :

2 sin 30° cos 30°

= $2\times\frac{1}{2} \times \frac{\sqrt{3}}{2}$

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

Right Hand Side :

sin 60° = $\frac{\sqrt{3}}{2}$

Hence, proved that 2 sin 30° cos 30° = sin 60°.

(iii) Left Hand Side :

2 sin 45° cos 45°

= $2\times\frac{1}{\sqrt2} \times \frac{1}{\sqrt2}$

= 1

Right Hand Side :

sin 90° = 1

Hence, proved that 2 sin 45° cos 45° = sin 90°.

(i) L.H.S. :

sin 2A = sin 2(45°) = sin 90°

= 1

R.H.S. :

2 sin A cos A = 2 sin 45° cos 45°

= $2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$

= 1.

Hence, proved that sin 2A = 2 sin A cos A.

(ii) Substituting value of A = 45° in cos 2A, we get :

cos 2A = cos 2(45°)

= cos 90°

= 0.

Substituting value of A = 45° in 2 cos²A - 1, we get :

⇒ 2 cos²A - 1

= 2 cos²45° - 1

= 2 (cos 45°)² - 1

= $2\Big(\frac{1}{\sqrt{2}}\Big)^2$ - 1

= $2 \times \frac{1}{2} - 1$

= 1 - 1

= 0.

Substituting value of A = 45° in 1 - 2 sin²A, we get :

⇒ 1 - 2 sin²A

= 1 - 2 sin²45°

= 1 - 2 (sin 45°)²

= 1 - $2\Big(\frac{1}{\sqrt{2}}\Big)^2$

= 1 - $2 \times \frac{1}{2}$

= 1 - 1

= 0.

Hence, proved that cos 2A = (2 cos²A - 1) = (1 - 2 sin²A).

(i) Left Hand Side:

sin 2A = sin 2(30°) = sin 60°

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

Right Hand Side:

$$\begin{aligned}\frac{2 \tan A}{1 + \tan^2A} = \frac{2\times \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}\\[1em] = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}\\[1em] = \frac{2}{\sqrt{3}}\times \frac{3}{4}\\[1em] = \frac{\sqrt{3}}{2}.\end{aligned}$$

Hence, proved that sin 2A = $\frac{2 \tan A}{1 + \tan^2A}$

(ii) Left Hand Side :

cos 2A = cos 2(30°) = cos 60°

= $\frac{1}{2}$

Right Hand Side :

$$\begin{aligned}\frac{1 - \tan^2A}{1 + \tan^2A} \\[1em] = \frac{1 - \tan^230°}{1 + \tan^230°} \\[1em] =\frac{1 - \Big(\frac{1}{\sqrt{3}}\Big)^2}{1 + \Big(\frac{1}{\sqrt{3}}\Big)^2}\\[1em] = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} \\[1em] = \frac{\frac{2}{3}}{\frac{4}{3}}\\[1em] = \frac{1}{2}.\end{aligned}$$

Hence, proved that cos 2A = $\Big(\frac{1 - \tan^2A}{1 + \tan^2A}\Big)$.

(i) Left Hand Side :

sin(A - B) = sin (45° - 45°) = sin 0

= 0

Right Hand Side :

sin A cos B - cos A sin B

= sin 45° cos 45° - cos 45° sin 45°

= $\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$

= 0.

Hence, proved that sin(A - B) = sin A cos B - cos A sin B.

(ii) Left Hand Side :

cos(A + B) = cos (45° + 45°) = cos 90°

= 0.

Right Hand Side :

cos A cos B - sin A sin B

= cos 45° cos 45° - sin 45° sin 45°

= $\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$

= $\frac{1}{2} - \frac{1}{2}$

= 0.

Hence, proved that cos(A + B) = cosA cosB - sin A sin B.

Left Hand Side :

(sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)²

= (sin 60° cos 30° + cos 60° sin 30°)² + (cos 60° cos 30° - sin 60° sin 30°)²

= $\Big(\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2} + \frac{1}{2}\times\frac{1}{2}\Big)^2 + \Big(\frac{1}{2}\times\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\times\frac{1}{2}\Big)^2$

= $\Big(\frac{3}{4} + \frac{1}{4})^2 + \Big(\frac{\sqrt{3}}{4}- \frac{\sqrt{3}}{4}\Big)^2$

= 1 + 0 = 1.

Right Hand Side = 1

Hence, proved that (sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)² = 1.

(i) Left Hand Side

sin (A + B) = sin (60° + 30°) = sin 90°

= 1

Right Hand Side

sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

= $\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2} + \frac{1}{2}\times\frac{1}{2}$

= $\frac{3}{4} + \frac{1}{4}$

= 1

Hence, proved that sin (A + B) = sin A cos B + cos A sin B.

(ii) Left Hand Side

cos (A + B) = cos (60° + 30°) = cos 90°

= 0

Right Hand Side

cos A cos B - sin A sin B

= cos 60° cos 30° - sin 60° sin 30°

= $\frac{1}{2}\times\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\times\frac{1}{2}$

= $\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}$

= 0.

Hence, proved that cos (A + B) = cos A cos B - sin A sin B.

(iii) Left Hand Side

cos (A - B) = cos (60° - 30°) = cos 30°

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

Right Hand Side

cos A cos B + sin A sin B

= cos 60° cos 30° + sin 60° sin 30°

$$\begin{aligned}\frac{1}{2}\times\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\times\frac{1}{2}\\[1em] = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}\\[1em] = \frac{2\sqrt{3}}{4}\\[1em] =\frac{\sqrt{3}}{2}\end{aligned}$$

Hence, proved that cos (A - B) = cos A cos B + sin A sin B.

(iv) Left Hand Side :

tan (A - B) = tan (60° - 30°) = tan 30°

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

Right Hand Side

$$\begin{aligned}\Rightarrow \Big(\frac{\tan A - \tan B}{1 + \tan A \tan B}\Big)\\[1em] = \Big(\frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)\\[1em] = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3}\times\frac{1}{\sqrt{3}}}\\[1em] =\frac{\frac{3-1}{\sqrt{3}}}{1 + 1}\\[1em] =\frac{\frac{2}{\sqrt{3}}}{2}\\[1em] = \frac{1}{\sqrt{3}}.\end{aligned}$$

Hence, proved that tan (A - B) = $\frac{\tan A - \tan B}{1 + \tan A \tan B}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A}\\[1em] = \frac{\cos 3(15^\circ) + 2 \cos 4(15^\circ)}{\sin 3(15^\circ) + 2\sin 4(15^\circ)}\\[1em] = \frac{\cos 45^\circ + 2 \cos 60^\circ}{\sin 45^\circ + 2\sin 60^\circ}\\[1em] = \frac{\frac{1}{\sqrt{2}}+ 2\times\frac{1}{2}}{\frac{1}{\sqrt{2}}+ 2\times\frac{\sqrt{3}}{2}}\\[1em] = \frac{\frac{1}{\sqrt{2}}+ 1 }{\frac{1}{\sqrt{2}}+ {\sqrt{3}}}\\[1em] = \frac{\frac{1 + \sqrt{2}}{\sqrt{2}}}{\frac{1 + \sqrt{6}}{\sqrt{2}}}\\[1em] = \frac{1 + \sqrt{2}}{1 + \sqrt{6}}.\end{aligned}$$

Hence, $\frac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A} = \frac{1 + \sqrt{2}}{1 + \sqrt{6}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{3\sin 3(20^\circ) + 2 \cos (2(20^\circ)+ 5^\circ)}{2 \cos 3(20^\circ) - \sin (2(20^\circ) - 10^\circ)}\\[1em] = \frac{3 \sin 60^\circ + 2 \cos 45^\circ}{2 \cos 60^\circ - \sin 30^\circ}\\[1em] = \frac{3\times\frac{\sqrt{3}}{2} + 2\times\frac{1}{\sqrt{2}}}{2\times\frac{1}{2} - \frac{1}{2}} \\[1em] = \frac{\frac{3\sqrt{3}}{2} + \frac{2}{\sqrt{2}}}{\frac{1}{{2}}}\\[1em] = \frac{\frac{3\sqrt{3}}{2} + {\sqrt{2}}}{\frac{1}{2}}\\[1em] = \frac{3\sqrt{3}+2\sqrt{2}}{2}\times\frac{2}{1}\\[1em] = {3\sqrt{3} + 2\sqrt{2}}.\end{aligned}$$

Hence, $\frac{3\sin 3A + 2\cos (2A + 5^\circ)}{2\cos 3A - \sin (2A - 10^\circ)} = {3\sqrt{3} + 2\sqrt{2}}.$

sin 30° = $\frac{1}{2}$

sin⁴30° = (sin 30°)⁴ = $\frac{1}{16}$

cos 60° = $\frac{1}{2}$

cos⁴60° = (cos 60°)⁴ = $\frac{1}{16}$

sin 45° = $\frac{1}{\sqrt{2}}$

cos²45° = (cos 45°)² = $\frac{1}{2}$

sin²90° = (sin 90°)² = 1

Left Hand Side

4(sin⁴30° + cos⁴60°)- 3(cos²45°- sin²90°)

= $4\Big(\frac{1}{16}+\frac{1}{16}\Big) - 3 \Big(\frac{1}{2}- 1\Big)$

= $4\Big(\frac{2}{16}\Big) - 3 \Big(\frac{-1}{2}\Big)$

= $\frac{1}{2} + \frac{3}{2}$

= 2.

Right Hand Side = 2

Hence, proved that 4(sin⁴30° + cos⁴60°) - 3(cos²45°- sin²90°) = 2.

(i) cos²30° = (cos 30°)² = $\Big(\frac{\sqrt{3}}{2}\Big)^2 = \frac{3}{4}$

sin²30° = (sin 30°)² = $\Big(\frac{1}{2}\Big)^2 = \frac{1}{4}$

Left Hand Side

cos 60° = $\frac{1}{2}$

Right Hand Side

cos²30° - sin²30°

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

Hence, proved that cos 60° = (cos²30° - sin²30).

(ii) Left Hand Side :

sin 60° = $\frac{\sqrt{3}}{2}$

Right Hand Side :

tan² 30° = (tan 30°)² = $\Big(\frac{1}{\sqrt{3}}\Big)^2 = \frac{1}{3}$

$$\begin{aligned}\Rightarrow \frac{2\tan 30^\circ}{1 + \tan^2 30^\circ}\\[1em] =\frac{2\times\frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}\\[1em] = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}\\[1em] = \frac{2}{\sqrt{3}}\times\frac{3}{4}\\[1em] = \frac{\sqrt{3}}{2}.\end{aligned}$$

Hence, proved that sin 60° = $\frac{2\tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{\sqrt{3}}{2}$.

(iii) Left Hand Side

cos 60° = $\frac{1}{2}$

Right Hand Side

$$\begin{aligned}\frac{1 -\tan^2 30^\circ}{1 + \tan^2 30^\circ}\\[1em] =\frac{1 -\frac{1}{3}}{1 + \frac{1}{{3}}}\\[1em] = \frac{\frac{{3} - 1}{{3}}}{\frac{{3} + 1}{{3}}}\\[1em] =\frac{2}{4} \\[1em] = \frac{1}{2}.\end{aligned}$$

Hence, proved that cos 60° = $\frac{1 -\tan^2 30^\circ}{1 + \tan^2 30^\circ} = \frac{1}{2}$.

sin 45° = cos 45° = $\frac{1}{\sqrt{2}}$.

x = 45 is the only value for which sin x° = cos x°.

(i) $\frac{\sin 27^\circ}{\cos 63^\circ} = \frac{\sin (90^\circ - 63^\circ)}{\cos 63^\circ}$

= $\frac{\cos 63^\circ}{\cos 63^\circ}$

= 1.

(ii) $\frac{\sec 38^\circ}{\cosec 52^\circ}$

= $\frac{\sec (90^\circ - 52^\circ)}{\cosec 52^\circ}$

= $\frac{\cosec 52^\circ}{\cosec 52^\circ}$

= 1.

(iii) $\frac{\tan 19^\circ}{\cot 71^\circ}$

= $\frac{\tan (90^\circ - 71^\circ)}{\cot 71^\circ}$

= $\frac{\cot 71^\circ}{\cot 71^\circ}$

= 1.

(i) $\frac{\sin 72^\circ}{\cos 18^\circ} - \frac{\sec 32^\circ}{\cosec 58^\circ}$

= $\frac{\sin (90^\circ - 18^\circ)}{\cos 18^\circ} - \frac{\sec (90^\circ - 58^\circ)}{\cosec 58^\circ}$

= $\frac{\cos 18^\circ}{\cos 18^\circ} - \frac{\cosec 58^\circ}{\cosec 58^\circ}$

= 1 - 1

= 0.

(ii) $\frac{2 \tan 53^\circ}{\cot 37^\circ} - \frac{\cot 80^\circ}{\tan 10^\circ}$

= $\frac{2\tan (90^\circ - 37^\circ)}{\cot 37^\circ} - \frac{\cot (90^\circ - 10^\circ)}{\tan 10^\circ}$

= $\frac{2 \cot 37^\circ}{\cot 37^\circ} - \frac{\tan 10^\circ}{\tan 10^\circ}$

= 2 - 1

= 1.

(i) $\frac{\sin 80^\circ}{\cos 10^\circ}$ + sin 59° sec 31°

= $\frac{\sin (90^\circ - 10^\circ)}{\cos 10^\circ}$ + sin (90° - 31°) sec 31°

= $\frac{\cos10^\circ}{\cos 10^\circ}$ + cos 31° sec 31°

= 1 + $\frac{1}{\sec 31^\circ}\times \sec 31^\circ$

= 1 + 1

= 2.

(ii) $\frac{\cos 35^\circ}{\sin 55^\circ} + \frac{\sin 11^\circ}{\cos 79^\circ}$ - cos 28° cosec 62°

= $\frac{\cos (90^\circ - 55^\circ)}{\sin 55^\circ} + \frac{\sin (90^\circ - 79^\circ)}{\cos 79^\circ}$ - cos (90°- 62°) cosec 62°

= $\frac{\sin 55^\circ}{\sin 55^\circ} + \frac{\cos 79^\circ}{\cos 79^\circ}$ - sin 62° cosec 62°

= 1 + 1 - $\frac{1}{\cosec 62^\circ}\times \cosec 62^\circ$

= 1 + 1 - 1

= 1.

(i) $\Big(\frac{\sin 39^\circ}{\cos 51^\circ}\Big)^2 + \Big(\frac{\cos 51^\circ}{\sin 39^\circ}\Big)^2$

= $\Big(\frac{\sin (90^\circ - 51^\circ)}{\cos 51^\circ}\Big)^2 + \Big(\frac{\cos (90^\circ - 39^\circ)}{\sin 39^\circ}\Big)^2$

= $\Big(\frac{\cos 51^\circ}{\cos 51^\circ}\Big)^2 + \Big(\frac{\sin 39^\circ}{\sin 39^\circ}\Big)^2$

= 1² + 1²

= 2.

(ii) $\Big(\frac{\cosec 34^\circ}{\sec 56^\circ}\Big)^2 + \Big(\frac{\cot 81^\circ}{\tan 9^\circ}\Big)^2$

= $\Big(\frac{\cosec (90^\circ -56^\circ)}{\sec 56^\circ}\Big)^2 + \Big(\frac{\cot (90^\circ - 9^\circ)}{\tan 9^\circ}\Big)^2$

= $\Big(\frac{\sec 56^\circ}{\sec 56^\circ}\Big)^2 + \Big(\frac{\tan 9^\circ}{\tan 9^\circ}\Big)^2$

= 1 + 1

= 2.

(i) 3 cos 80° cosec 10° + 2 cos 59° cosec 31°

= 3 cos (90° - 10°) cosec 10° + 2 cos (90° - 31°) cosec 31°

= 3 sin 10° cosec 10° + 2 sin 31°cosec 31°

= $3\times\frac{1}{\cosec 10^\circ}\times \cosec 10^\circ + 2\times \frac{1}{\cosec 31^\circ}\times \cosec 31^\circ$

= 3 + 2

= 5.

(ii) 5 sin 70° sec 20° - 3 sin 50° sec 40°

= 5 sin (90° - 20°) sec 20° - 3 sin (90° - 40°) sec 40°

= 5 cos 20° sec 20° - 3 cos 40° sec 40°

= $5\times\frac{1}{\sec 20^\circ}\times \sec 20^\circ - 3\times \frac{1}{\sec 40^\circ}\times \sec 40^\circ$

= 5 - 3

= 2.

(i) Solving,

$$\begin{aligned}\frac{\sin 35^\circ \cos 55^\circ + \cos 35^\circ \sin 55^\circ}{\cosec^2 10^\circ - \tan^2 80^\circ}\\[1em] = \frac{\sin (90^\circ - 55^\circ) \cos 55^\circ + \cos (90^\circ - 55^\circ) \sin 55^\circ}{\cosec^2 (90^\circ - 80^\circ) - \tan^2 80^\circ}\\[1em] = \frac{\cos 55^\circ \cos 55^\circ + \sin 55^\circ \sin 55^\circ}{\sec^2 80^\circ - \tan^2 80^\circ}\\[1em] = \frac{\cos^2 55^\circ + \sin^2 55^\circ}{1} \\[1em] = 1.\end{aligned}$$

Hence, $\frac{\sin 35^\circ \cos 55^\circ + \cos 35^\circ \sin 55^\circ}{\cosec^2 10^\circ - \tan^2 80^\circ} = 1.$

(ii) Solving,

$$\begin{aligned}\Rightarrow 4 \times \tan 60^\circ \sec 30^\circ + \frac{\sin 31^\circ \sec 59^\circ + \cot 59^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \tan^2 45^\circ}\\[1em] \Rightarrow 4 \times\frac{\sin 60^\circ}{\cos 60^\circ} \times \frac{1}{\cos 30^\circ} + \frac{\sin (90^\circ - 59^\circ) \frac{1}{\cos 59^\circ }+ \cot (90^\circ - 31^\circ) \cot 31^\circ}{8 \sin^2 30^\circ - \frac{\sin^2 45^\circ}{\cos^2 45^\circ}}\\[1em] \Rightarrow 4 \times\frac{\sin (90^\circ -30^\circ)}{\cos 60^\circ} \times \frac{1}{\cos 30^\circ} + \frac{\cos 59^\circ \times \frac{1}{\cos 59^\circ }+ \tan 31^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \frac{\sin^2 (90^\circ - 45^\circ)}{\cos^2 45^\circ}}\\[1em] \Rightarrow 4 \times\frac{\cos 30^\circ}{\cos 60^\circ} \times \frac{1}{cos 30^\circ} + \frac{1 + 1}{8 \sin^2 30^\circ - \frac{\cos^2 45^\circ}{\cos^2 45^\circ}}\\[1em] \Rightarrow 4 \times\frac{1}{\cos 60^\circ} + \frac{1 + 1}{8 \sin^2 30^\circ - 1}\\[1em] \Rightarrow 4 \times \sec 60^\circ + \frac{2}{8 \times \frac{1}{4} - 1}\\[1em] \Rightarrow 4 \times 2 + \frac{2}{2 - 1}\\[1em] \Rightarrow 8 + 2 \\[1em] \Rightarrow 10.\end{aligned}$$

Hence, 4 tan 60° sec 30° + $\frac{\sin 31^\circ \sec 59^\circ + \cot 59^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \tan^2 45^\circ}$ = 10.

(i) $\frac{\sin 29^\circ 31'}{\cos 60^\circ 29'}$

= $\frac{\sin (90^\circ - 60^\circ 29')}{\cos 60^\circ 29'}$

= $\frac{\cos 60^\circ 29'}{\cos 60^\circ 29'}$

= 1.

(ii) $\frac{\tan 20^\circ 43'}{\cot 69^\circ 17'}$

= $\frac{\tan (90^\circ - 69^\circ 17')}{\cot 69^\circ 17'}$

= $\frac{\cot 69^\circ 17'}{\cot 69^\circ 17'}$

= 1.

(iii) $\frac{\sin 28^\circ 54'}{\sec 61^\circ 6'} + \frac{\cos 28^\circ 54'}{\cosec 61^\circ 6'}$

= $\frac{\sin (90^\circ - 61^\circ 6')}{\sec 61^\circ 6'} + \frac{\cos (90^\circ - 61^\circ 6')}{\cosec 61^\circ 6'}$

= $\frac{\cos 61^\circ 6'}{\sec 61^\circ 6'} + \frac{\sin 61^\circ 6'}{\cosec 61^\circ 6'}$

= cos 61°6' cos 61°6' + sin 61°6' sin 61°6'

= cos² 61°6' + sin² 61°6'

= 1.

(i) sin 37° - cos 53°

= sin(90° - 53°) - cos 53°

= cos 53° - cos 53°

= 0.

(ii) cos² 25° - sin² 65°

= cos² (90° -65°) - sin² 65°

= sin² 65° - sin² 65°

= 0.

(iii) sin² 23° + sin² 67°

= sin² (90° - 67°) + sin² 67°

= cos² 67° + sin² 67°

= 1.

(i) sec² 36° - cot² 54°

= sec² (90° - 54°) - cot² 54°

= cosec² 54° - cot² 54°

= 1.

(ii) cosec² 38° - tan² 52°

= cosec² (90° - 52°) - tan² 52°

= sec² 52° - tan² 52°

= 1.

(iii) cos² 24° + cos² 66°

= cos² (90° - 66°) + cos² 66°

= sin² 66° + cos² 66°

= 1.

(i) $\frac{\cos^2 34^\circ + \cos^2 56^\circ}{\sin^2 59^\circ + \sin^2 31^\circ}$

= $\frac{\cos^2 (90^\circ - 56^\circ) + \cos^2 56^\circ}{\sin^2 (90^\circ - 31^\circ) + \sin^2 31^\circ}$

= $\frac{\sin^2 56^\circ + \cos^2 56^\circ}{\cos^2 31^\circ + \sin^2 31^\circ}$

= 1.

(ii) $\frac{\cosec^2 67^\circ - \tan^2 23^\circ}{\sec^2 20^\circ - \cot^2 70^\circ}$

= $\frac{\cosec^2 (90^\circ - 23^\circ) - \tan^2 23^\circ}{\sec^2 (90^\circ - 70^\circ) - \cot^2 70^\circ}$

= $\frac{\sec^2 23^\circ - \tan^2 23^\circ}{\cosec^2 70^\circ - \cot^2 70^\circ}$

= 1.

(i) sin 73° cos 17° + cos 73° sin 17° = 1

Solving L.H.S.,

sin 73° cos 17° + cos 73° sin 17°

= sin (90° - 17°) cos 17° + cos (90° - 17°) sin 17°

= cos 17° cos 17° + sin 17° sin 17°

= cos² 17° + sin² 17°

= 1.

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

Hence, proved that sin 73° cos 17° + cos 73° sin 17° = 1.

(ii) sin 40° sec 50° + cos 40° cosec 50° = 2

Solving L.H.S.

sin (90° - 50°) sec 50° + cos (90° - 50°) cosec 50°

= cos 50° sec 50° + sin 50° cosec 50°

= $\frac{1}{\sec 50^\circ}\times \sec 50^\circ + \frac{1}{\cosec 50^\circ}\times \cosec 50^\circ$

= 1 + 1 = 2

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

Hence, proved that sin 40° sec 50° + cos 40° cosec 50° = 2.

(iii) sec²75° - cot²15° = 1

Solving L.H.S.

= sec²75° - cot²15°

= sec²(90° - 15°) - cot²15°

= cosec²15° - cot²15°

= 1

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

Hence, proved that sec²75° - cot²15° = 1.

(iv) cos²23° + cos²67° = 1

Solving L.H.S.

cos²23° + cos²67°

= cos²(90° - 67°) + cos²67°

= sin²67° + cos²67°

= 1

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

Hence, proved that cos²23° + cos²67° = 1.

(v) cosec²56° - tan²34° = 1

Solving L.H.S.

cosec²56° - tan²34°

= cosec²(90° - 34°) - tan²34°

= sec²34° - tan²34°

= 1

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

Hence, proved that cosec²56° - tan²34° = 1.

(vi) tan 10° tan 15° tan 75° tan 80° = 1

Solving L.H.S.

tan 10° tan 15° tan 75° tan 80°

= tan (90° - 80°) tan (90°- 75°) tan 75° tan 80°

= cot 80° cot 75° tan 75° tan 80°

= $\frac{1}{\tan80^\circ}\times \tan 80^\circ \times \frac{1}{\tan 75^\circ}\times \tan 75^\circ$

= 1

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

Hence, proved that tan 10° tan 15° tan 75° tan 80° = 1.

(i) cos 81° + cot 81°

= cos (90° - 9°) + cot (90° - 9°)

= sin 9° + tan 9°

Hence, cos 81° + cot 81° = sin 9° + tan 9°.

(ii) cos 76° + sec 76°

= cos (90° - 14°) + sec (90° - 14°)

= sin 14° + cosec 14°

Hence, cos 76° + sec 76° = sin 14° + cosec 14°.

(iii) sec 63° + cosec 49°

= sec (90° - 27°) + cosec (90° - 41°)

= cosec 27° + sec 41°

Hence, sec 63° + cosec 49° = cosec 27° + sec 41°.

(iv) sin 59° + cos 56°

= sin (90° - 31°) + cos (90° - 34°)

= cos 31° + sin 34°

Hence, sin 59° + cos 56° = cos 31° + sin 34°.

As,

cos θ = sin (90° - θ)

Solving,

cos(65° + θ) - sin (25° - θ)

= sin (90° - (65° + θ)) - sin(25° - θ)

= sin (90° - 65° - θ) - sin(25° - θ)

= sin (25° - θ) - sin (25° - θ)

= 0.

Hence, proved that cos(65° + θ) - sin (25° - θ) = 0.

Solving,

sin (50° + θ) - cos (40° - θ)

= cos (90° - (50° + θ)) - cos (40° - θ)

= cos (90° - 50° - θ) - cos (40° - θ)

= cos (40° - θ) - cos (40° - θ)

= 0.

Hence, proved that sin (50° + θ) - cos (40° - θ) = 0.

Solving,

tan(45° - A) tan(45° + A)

= tan[90° - (45° + A)] tan(45° + A)

= cot(45° + A) tan(45° + A)

= $\frac{1}{\tan(45^\circ + A)}\times \tan(45^\circ + A)$

= 1.

Hence, proved that tan(45° - A) tan(45° + A) = 1.

(i) $\frac{\cos A}{\sin(90^\circ - A)} + \frac{\sin A}{\cos (90^\circ - A)}$ = 2

Solving L.H.S.,

$\frac{\cos A}{\sin(90^\circ - A)} + \frac{\sin A}{\cos (90^\circ - A)}$

= $\frac{\cos A}{\cos A} + \frac{\sin A}{\sin A}$

= 1 + 1

= 2.

Hence, proved that $\frac{\cos A}{\sin(90^\circ - A)} + \frac{\sin A}{\cos (90^\circ - A)}$ = 2.

(ii) $\frac{\sin A}{\sin(90^\circ - A)} + \frac{\cos A}{\cos (90^\circ - A)}$ = sec(90° - A) cosec(90° - A)

Solving L.H.S.,

$$\begin{aligned}\Rightarrow\frac{\sin A}{\sin(90^\circ - A)} + \frac{\cos A}{\cos (90^\circ - A)}\\[1em] = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \\[1em] = \frac{\sin^2 A + \cos^2 A}{\cos A \sin A} \\[1em] = \frac{1}{\text{sin A cos A}}\\[1em] = \cosec A \sec A\\[1em] = \sec(90^\circ - A) \cosec(90^\circ - A).\end{aligned}$$

Hence, proved that $\frac{\sin A}{\sin(90^\circ - A)} + \frac{\cos A}{\cos (90^\circ - A)}$ = sec(90° - A) cosec(90° - A).

(iii) sin(90° - A) cos(90° - A) = $\frac{\tan A}{1 + \tan^2 A}$

Solving L.H.S.,

sin(90° - A) cos(90° - A)

= cos A sin A

Solving R.H.S.,

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

Since, L.H.S. = R.H.S. = sin A. cos A

Hence, proved that sin(90° - A) cos(90° - A) = $\frac{\tan A}{1 + \tan^2 A}$.

Let A be the position of balloon and C be the position of meterological station. Then length of cable (AC) = 200 m.

Let height of the ballon from the ground = x

In triangle ABC,

sin 60° = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

$\frac{\sqrt{3}}{2} = \frac{x}{200}$

x = $\frac{200 \sqrt{3}}{2}$

x = $100\sqrt{3}$

x = 100(1.73)

x = 173 m.

Hence, height of the balloon from the ground = 173 m.

Let AB be the wall and AC be the ladder.

In triangle ABC,

$$\begin{aligned}\Rightarrow \cos 60^\circ = \frac{BC}{AC} \\[1em] \Rightarrow \frac{1}{2} = \frac{9.5}{AC} \\[1em] \Rightarrow AC = 2 \times 9.5 = 19 \text{ m}.\end{aligned}$$

Hence, length of ladder = 19 m.

Let position of kite be A and C be the point on ground.

AC = length of string = 150 m.

AB = Height of the kite above the ground

sin 60° = $\frac{AB}{AC}$

$\frac{\sqrt{3}}{2} = \frac{AB}{150}$

AB = $75\sqrt{3} = 75 \times 1.73$ = 129.75 m

Hence, height of the kite above ground = 129.75 m.

Let A be the position of kite and C be the point from where string is attached.

AB = 120 m

In triangle ABC,

$$\begin{aligned}\Rightarrow \sin 60^\circ = \frac{AB}{AC} \\[1em] \Rightarrow \frac{\sqrt{3}}{2} = \frac{120}{AC} \\[1em] \Rightarrow AC = \frac{120 \times 2}{\sqrt{3}} \\[1em] \Rightarrow AC = \frac{240}{\sqrt{3}} \\[1em] \Rightarrow AC = \frac{240 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow AC = \frac{240\sqrt{3}}{3} \\[1em] \Rightarrow AC = 80\sqrt{3} = 138.4 \text{ m}.\end{aligned}$$

Hence, length of the string = 138.4 m.

Given,

∠A = 30°, BC = 8 cm

sin 30° = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

$\frac{1}{2} = \frac{BC}{AC}$

$\frac{1}{2} = \frac{8}{AC}$

AC = 16 cm

Now,

cos 30° = $\frac{\text{Base}}{\text{Hypotenuse}}$

$\frac{\sqrt{3}}{2} = \frac{AB}{AC}$

$\frac{\sqrt{3}}{2} = \frac{AB}{16}$

AB = $8\sqrt{3}$ cm

By angle sum property of triangle,

∠A + ∠B + ∠C = 180°

30° + 90° + ∠C = 180°

∠C + 120° = 180°

∠C = 180° - 120° = 60°

Hence, ∠C = 60°, AC = 16 cm and AB = $8\sqrt{3}$ cm.

In △BAC,

Base = AB = 12 cm and ∠BAC = 30°

Perpendicular = BC

tan 30° = $\frac{BC}{AB}$

$\frac{1}{\sqrt{3}} = \frac{BC}{12}$

BC = $\frac{12}{\sqrt{3}} = 4\sqrt{3}$ cm

Now,

cos 30° = $\frac{AB}{AC}$

$\frac{\sqrt{3}}{2} = \frac{12}{AC}$

AC = $\frac{24}{\sqrt{3}} = 8\sqrt{3}$ cm.

Hence, BC = $4\sqrt{3}$ cm and AC = $8\sqrt{3}$ cm.

Given,

sin(θ - 15°)= $\frac{1}{2}$

sin(θ - 15°) = sin 30°

θ - 15° = 30°

θ = 30° + 15° = 45°

so,

cos(θ - 15°) = cos(45° - 15°) = cos 30° = $\frac{\sqrt{3}}{2}$

Hence, option 2 is the correct option.

Given,

cos(θ - 30°) = $\frac{1}{2}$

cos(θ - 30°) = cos 60°

θ - 30° = 60°

θ = 60° + 30° = 90°

Then,

tan θ = tan 90° = undefined.

Hence, option 4 is the correct option.

Given,

x tan 30° = cos 60°

$x \times \frac{1}{\sqrt{3}} = \frac{1}{2}$

x = $\frac{\sqrt{3}}{2}$

Hence, option 3 is the correct option.

tan (θ + 15°) = 1

tan (θ + 15°) = tan 45°

θ + 15° = 45°

θ = 45° - 15°= 30°

Then

cos 2θ = cos 2(30°) = cos 60° = $\frac{1}{2}$.

Hence, option 1 is the correct option.

Given,

sin θ = cos θ

This is possible in case of θ = 45° as sin 45° = cos 45° = $\frac{1}{\sqrt{2}}$.

θ = 45°

Then,

sec (θ + 15°) = sec (45° + 15°)

= sec 60°

= 2.

Hence, option 2 is the correct option.

Given,

cos 2θ = 0

cos 2θ = cos 90°

2θ = 90°

θ = 45°

Then,

cot(θ - 15°) = cot (45° - 15°) = cot 30° = $\sqrt{3}$.

Hence, option 3 is the correct option.

Given,

sin(θ + 18°) = $\frac{1}{2}$

sin(θ + 18°) = sin 30°

θ + 18° = 30°

θ = 30° - 18° = 12°

Then,

cosec 5θ = cosec 5(12°) = cosec 60° = $\frac{2}{\sqrt{3}}$.

Hence, option 4 is the correct option.

sin θ = $\frac{8}{17}$

sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

Let Perpendicular = 8x and Hypotenuse = 17x

We will find Base by using Pythagoras Theorem,

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = (17x)² - (8x)²

Base² = 289x² - 64x²

Base² = 225x²

Base = 15x

Then,

cot θ = $\frac{\text{Base}}{\text{Perpendicular}}$

= $\frac{15x}{8x} = \frac{15}{8}$.

Hence, option 1 is the correct option.

sin θ = $\frac{1}{2}$

sin 30° = $\frac{1}{2}$

sin θ = sin 30°

θ = 30°

Then,

3cos θ - 4 cos³ θ = 3 cos 30° - 4 (cos 30°)³

= $3\times \frac{\sqrt{3}}{2}- 4 \times \Big(\frac{\sqrt{3}}{2}\Big)^3$

= $\frac{3\sqrt{3}}{2}- \frac{12\sqrt{3}}{8}$

= $\frac{3\sqrt{3}}{2}- \frac{3\sqrt{3}}{2}$

= 0.

Hence, option 1 is the correct option.

Given,

5 cot θ = 3

cot θ = $\frac{3}{5}$

cot θ = $\frac{\cos θ}{\sin θ}$

Given,

$\frac{(5 \sin θ - 3 \cos θ)}{(4 \sin θ + 3 \cos θ)}$

Dividing above equation by sin θ, we get :

$$\begin{aligned}\Rightarrow \frac{\frac{5\sin θ - 3\cos θ}{\sin θ}}{\frac{4\sin θ + 3 \cos θ}{\sin θ}} \\[1em] \Rightarrow \frac{(5 - 3 \cot θ)}{(4 + 3 \cot θ)}\\[1em] = \frac{\Big(5 - 3\times \frac{3}{5}\Big)}{\Big(4 + 3\times \frac{3}{5}\Big)}\\[1em] = \frac{\Big(5 - \frac{9}{5}\Big)}{\Big(4 + \frac{9}{5}\Big)}\\[1em] = \frac{\Big(\frac{{25 -9}}{5}\Big)}{\Big( \frac{20 + 9}{5}\Big)}\\[1em] = \frac{\Big(\frac{{16}}{5}\Big)}{\Big( \frac{29}{5}\Big)}\\[1em] = \frac{16}{5}\times \frac{5}{29}\\[1em] = \frac{16}{29}.\end{aligned}$$

Hence, option 2 is the correct option.

Perpendicular = AB = 5 cm

Base = BC = 12 cm

By using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

Hypotenuse² = (5)² + (12)²

Hypotenuse² = 25² + 144²

Hypotenuse² = 169²

Hypotenuse = 13 cm

AC = 13 cm

sin C = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{5}{13}$.

Hence, option 2 is the correct option.

Solving,

⇒ sin θ cos (90° - θ) + cos θ sin(90° - θ)

⇒ sin θ sin θ + cos θ cos θ

⇒ sin²θ + cos²θ

⇒ 1.

Hence, option 2 is the correct option.

⇒ sin²25° + sin²65°

⇒ sin²25° + sin²(90° - 25°)

⇒ sin²25° + cos²25°

⇒ 1.

Hence, option 4 is the correct option.

Given,

$\Rightarrow \frac{\tan 35^\circ}{\cot 55^\circ} + \frac{\cot 78^\circ}{\tan 12^\circ}$

= $\frac{\tan 35^\circ}{\cot (90^\circ - 35^\circ)} + \frac{\cot (90^\circ - 12^\circ)}{\tan 12^\circ}$

= $\frac{\tan 35^\circ}{\tan 35^\circ} + \frac{\tan 12^\circ}{\tan 12^\circ}$

= 1 + 1 = 2.

Hence, option 2 is the correct option.

3 sin θ + 4 cos θ = 5

4 cos θ = 5 - 3 sin θ

Squaring Both Sides,

(4 cos θ)² = (5 - 3 sin θ)²

16 cos²θ = 25 + 9 sin²θ - 30 sin θ

Putting cos²θ = 1 - sin²θ

16 ( 1 - sin²θ) = 25 + 9 sin²θ - 30 sin θ

16 - 16 sin²θ = 25 + 9 sin²θ - 30 sin θ

25 + 9 sin²θ - 30 sin θ - 16 + 16 sin²θ = 0

25 sin²θ - 30 sin θ + 9 = 0

25 sin²θ - 15 sin θ - 15 sin θ + 9 = 0

5 sin θ(5sin θ - 3) - 3(5sin θ - 3) = 0

(5 sin θ - 3)(5sin θ - 3)= 0

(5 sin θ - 3)² = 0

5 sin θ - 3 = 0

$\Rightarrow \sin θ = \frac{3}{5}$

Hence, option 2 is the correct option.

Solving,

⇒ tan 5° tan 25° tan 30° tan 65° tan 85°

⇒ tan 5° tan 85° tan 25° tan 65° tan 30°

⇒ tan 5° tan (90° - 5°) tan 25° tan (90° - 25°) tan 30°

⇒ tan 5° cot 5° tan 25° cot 25° tan 30°

⇒ $1 \times 1 \times$ tan 30°

⇒ $\frac{1}{\sqrt{3}}$.

Hence, option 3 is the correct option.

(cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)

$$\begin{aligned}\Rightarrow \Big(1 + \frac{1}{\sqrt{2}} + \frac{1}{2}\Big) \Big(1 - \frac{1}{\sqrt{2}} + \frac{1}{2}\Big)\\[1em] = \big(\frac{3}{2} + \frac{1}{\sqrt{2}}\Big) \Big(\frac{3}{2} - \frac{1}{\sqrt{2}}\Big)\\[1em] = \Big(\frac{3}{2}\Big)^2 - \Big(\frac{1}{\sqrt{2}}\Big)^2\\[1em] = \frac{9}{4} - \frac{1}{2}\\[1em] = \frac{9 - 2}{4} = \frac{7}{4}\end{aligned}$$

Hence, option 3 is the correct option.

1. Perpendicular = BC = $2\sqrt{3}$ m and Base = AB = 6 m

Hypotenuse = AC

We will find Hypotenuse by using Pythagoras Theorem,

Hypotenuse² = Perpendicular² + Base²

Hypotenuse² = $(2\sqrt{3})^2$ + 6²

Hypotenuse² = 12 + 36

Hypotenuse² = 48

Hypotenuse = $\sqrt{48} = 4\sqrt{3}$ m

Hence, option (d) is the correct option.

2. sin A = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

sin A = $\frac{BC}{AC} = \frac{2\sqrt{3}}{4\sqrt{3}}$

sin A = $\frac{1}{2}$

sin A = sin 30°

A = 30°

Hence, option (a) is the correct option.

3. sin C = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

sin C = $\frac{AB}{AC} = \frac{6}{4\sqrt{3}}$

sin C = $\frac{\sqrt{3}}{2}$

sin C = sin 60°

C = 60°

Hence, option (c) is the correct option.

4. As, A = 30°

cos 2(30°) = cos 60° = $\frac{1}{2}$

Hence, option (b) is the correct option.

5. As, C = 60°

$2\sin(\frac{C}{2}) = 2\sin(\frac{60}{2})$

= 2 sin 30°

= $2 \times \frac{1}{2}$

= 1.

Hence, option (a) is the correct option.

1. AB = Perpendicular = 4 inches

BC = Hypotenuse = 8 inches

sin∠ACB = $\frac{\text Perpendicular}{\text Hypotenuse}$

sin∠ACB = $\frac{4}{8} = \frac{1}{2}$

sin∠ACB = sin 30°

∠ACB = 30°

Hence, option (a) is the correct option.

2. ∠ABC + ∠BAC + ∠ACB = 180°

∠ABC + 90° + 30° = 180°

∠ABC = 180° - 120° = 60°

tan ∠ABC = tan 60° = $\sqrt{3}$.

Hence, option (b) is the correct option.

3. sin²∠ACB + sin²∠ABC

= sin²30° + sin²60°

= $\Big(\frac{1}{2}\Big)^2 + \Big(\frac{\sqrt{3}}{2}\Big)^2$

= $\frac{1}{4} + \frac{3}{4}$

= $\frac{4}{4}$

= 1.

Hence, option (c) is the correct option.

4. AC = Base

By pythagoras theorem,

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

AC² = BC² - AB²

AC² = 8² - 4²

AC² = 64 - 16

AC² = 48

AC = $\sqrt{48}$

AC = $4\sqrt{3}$ inches.

Hence, option (d) is the correct option.

5. Solving,

sin ∠ACB + cos ∠ABC + cot ∠BAC

= sin 30° + cos 60° + cot 90°

= $\frac{1}{2} + \frac{1}{2} + 0$

= 1.

Hence, option (b) is the correct option.

Given,

$$\begin{aligned}\Rightarrow \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ\\[1em] \Rightarrow \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}\\[1em] \Rightarrow \frac{3}{4} + \frac{1}{4}\\[1em] \Rightarrow \frac{4}{4}\\[1em] = 1.\end{aligned}$$

Assertion (A) is false.

As,

sin 90° = 1 and sin 0° = 0.

So,

Reason (R) is false.

Hence, Option 4 is the correct option.

In a right angled △ABC,

∠ABC = 90°, AB = 3 cm, BC = 4 cm

By pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

AC² = (BC)² + (AB)²

AC² = 4² + 3²

AC² = 16 + 9

AC² = 25

AC = $\sqrt{25}$

AC = 5 cm.

$$\begin{aligned}\Rightarrow \sin A = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{4}{5}\\[1em] \Rightarrow \cos C = \frac{\text{base}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{4}{5}\\[1em] \Rightarrow \sin A = \cos C\end{aligned}$$

Assertion (A) is true.

As, $\frac{\sin θ}{\cos θ}$ = tan θ

but sin θ $\times$ cos θ is not equal to cot θ.

So, Reason (R) is false.

Hence, Option 1 is the correct option.

$$\begin{aligned}\Rightarrow \frac{\sin 27^\circ}{\cos 63^\circ}\\[1em] = \frac{\sin (90^\circ - 63^\circ)}{\cos 63^\circ}\\[1em] = \frac{\cos 63^\circ}{\cos 63^\circ}\\[1em] = 1\end{aligned}$$

Assertion (A) is true.

As, sin(90° - θ) = cos θ

and

cos(90° - θ) = sin θ

So,

Reason (R) is true.

Hence, Option 3 is the correct option.