Trigonometrical Ratios — Exercise 20(A)

Look at the figures given below :

From these figures, write down the values of :

(i) sin x

(ii) tan x

(iii) sec x

(iv) cos y

(v) cot y

(vi) cosec y

(vii) sin z

(viii) cos z

(ix) tan z

Answer

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

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

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

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

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

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

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

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

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

In the given figure, ∠B = 90°, AB = 4 units and BC = 3 units. Find:

(i) sin A

(ii) cos A

(iii) cot A

(iv) sin C

(v) sec C

(vi) tan C

Answer

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{3}{5}$

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

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

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

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

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

From the given figure, write down the values of :

(i) sin B

(ii) tan B

(iii) cos C

(iv) cot C

(v) (sin B cos C + cos B sin C)

(vi) (sec² C - tan² C)

Answer

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AC}{BC} = \dfrac{8}{17}$

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

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

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

(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 = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{AB}{BC} = \dfrac{15}{17}$

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

Substituting the values, we get :

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

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 = $\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{BC}{AC} = \dfrac{17}{8}$

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

Now putting the values of sec C & tan C

sec²C - tan²C

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

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

= $\dfrac{64}{64}$

= 1.

Hence, sec²C - tan²C = 1.

In the given figure, AD ⊥ BC.

If AB = 13 cm, BD = 5 cm and DC = 16 cm, find the values of :

(i) sin B

(ii) sec B

(iii) cot B

(iv) cos C

(v) cosec C

(vi) tan C

Answer

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AD}{AB} = \dfrac{12}{13}$

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

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

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

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

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

If sin θ = $\dfrac{1}{\sqrt{2}}$, find the values of other trigonometrical ratios for θ.

Answer

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

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 = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{x}{\sqrt{2}x} = \dfrac{1}{\sqrt{2}}$

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

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

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

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

If tan θ = $\dfrac{8}{15}$, find the values of other trigonometrical ratios for θ.

Answer

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

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 = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{8x}{17x} = \dfrac{8}{17}$.

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

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

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

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

If cosec θ = $\sqrt{10}$, find the values of other trigonometrical ratios for θ.

Answer

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

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{x}{\sqrt{10}x} = \dfrac{1}{\sqrt{10}}$

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

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

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

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

If sin θ = $\dfrac{3}{5}$ and θ is an acute angle, find the values of cos θ and tan θ.

Answer

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

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 = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{4x}{5x} = \dfrac{4}{5}$

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

If tan θ = $\dfrac{5}{12}$ and θ is acute, find the values of sin θ and cos θ.

Answer

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

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{5x}{13x} = \dfrac{5}{13}$

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

If sin θ = $\dfrac{\sqrt{3}}{2}$, find the value of (cosec θ + cot θ).

Answer

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

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 = $\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{2x}{\sqrt{3}x} = \dfrac{2}{\sqrt{3}}$

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

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

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

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

Hence, cosec θ + cot θ = $\sqrt{3}$.

If 13 sin θ = 5, find the value of

$\dfrac{\text{5 sin θ - 2 cos θ}}{\text{tan θ}}$

Answer

13 sin θ = 5

sin θ = $\dfrac{5}{13}$

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

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 = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{12x}{13x} = \dfrac{12}{13}$

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

Substituting values, we get :

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

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

If cot θ = $\dfrac{1}{\sqrt{3}}$, show that $\Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{3}{5}$.

Answer

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

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{\sqrt{3}x}{2x} = \dfrac{\sqrt{3}}{2}$

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

Substituting values we get :

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

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

If sec θ = $\dfrac{13}{5}$, show that $\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}}$ = 3.

Answer

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

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{12x}{13x} = \dfrac{12}{13}$

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

Substituting values we get :

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

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

If 3 tan θ = 4, show that $\Big(\dfrac{3\sin θ + 2\cos θ}{3\sin θ - 2\cos θ}\Big)$= 3.

Answer

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

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{4x}{5x} = \dfrac{4}{5}$

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

Substituting values we get :

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

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

If cot θ = $\dfrac{q}{p}$, show that $\Big(\dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}\Big)= \dfrac{p^2 - q^2}{p^2 + q^2}$.

Answer

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

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

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

Substituting values we get :

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

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

If 4 cot θ = 3, show that $\Big(\dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ}\Big) = \dfrac{1}{7}$.

Answer

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

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{4x}{5x} = \dfrac{4}{5}$

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

Substituting values we get :

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

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

Use the adjoining figure and write the values of :

(i) sin x°

(ii) cos y°

(iii) 3 tan x° - 2 sin y° + 4 cos y°

Answer

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° = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{8}{17}$.

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

(iii) 3 tan x° - 2 sin y° + 4 cos y°

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

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

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

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

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

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

Using the adjoining figure, calculate the values of :

(i) cos θ

(ii) tan Φ

(iii) cosec Φ

Answer

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 = $\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{AB}{AC} = \dfrac{12}{13}$.

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

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

If (tan θ + cot θ) = 5, find the value of (tan²θ + cot²θ).

Answer

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 + $2\tan \theta \times \dfrac{1}{\tan \theta}$ = 25

⇒ tan²θ + cot²θ + 2 = 25

⇒ tan²θ + cot²θ = 25 - 2

⇒ tan²θ + cot²θ = 23.

Hence, tan²θ + cot²θ = 23.

If (cos θ + sec θ ) = $\dfrac{5}{2}$, find the value of (cos²θ + sec²θ).

Answer

Given,

(cos θ + sec θ ) = $\dfrac{5}{2}$

Squaring both sides,

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

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

cos²θ + sec²θ + 2 = $\dfrac{25}{4}$

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

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

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

Evaluate x and y from the given figure.

Answer

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° = $\dfrac{\text{base}}{\text{hypotenuse}}$

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

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

x = 5 m.

In △BDC,

BC = $5\sqrt{2}$ m

CD = x = 5 m

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

sin y° = $\dfrac{1}{\sqrt{2}}$

sin y° = sin 45°

y° = 45°.

Hence, x = 5 m and y° = 45°.

In the given figure, △ABC is right angled at B.

If AC = 20 cm and tan A = $\dfrac{3}{4}$, find the lengths of AB and BC.

Answer

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

Given,

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

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.

If cos θ = $\dfrac{2x}{1 + x^2}$, find the values of sin θ and tan θ in terms of x.

Answer

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

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 = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{x^2 - 1}{1 + x^2}$

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