Trigonometrical Ratios of Standard Angles — Exercise 17.1

Find the values of :

(i) 7 sin 30° cos 60°

(ii) 3 sin² 45° + 2 cos² 60°

(iii) cos² 45° + sin² 60° + sin² 30°

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

Answer

(i) 7 sin 30° cos 60°

$= 7 \times \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{7}{4}.$

Hence, 7 sin 30° cos 60° = $\dfrac{7}{4}$.

(ii) 3 sin² 45° + 2 cos² 60°

$= 3 \times \Big(\dfrac{1}{\sqrt{2}}\Big)^2 + 2 \times \Big(\dfrac{1}{2}\Big)^2 \\[1em] = 3 \times \dfrac{1}{2} + 2 \times \dfrac{1}{4} \\[1em] = \dfrac{3}{2} + \dfrac{1}{2} \\[1em] = \dfrac{4}{2} \\[1em] = 2.$

Hence, 3 sin² 45° + 2 cos² 60° = 2.

(iii) cos² 45° + sin² 60° + sin² 30°

$= \Big(\dfrac{1}{\sqrt{2}}\Big)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{1}{2} + \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{2 + 3 + 1}{4} \\[1em] = \dfrac{6}{4} \\[1em] = \dfrac{3}{2}.$

Hence, cos² 45° + sin² 60° + sin² 30° = $\dfrac{3}{2}$.

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

$= 0 + \Big(\dfrac{1}{\sqrt{2}}\Big)^2 \times \dfrac{1}{2} \times 1 \\[1em] = 0 + \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{1}{4}.$

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

Find the values of

(i) $\dfrac{\text{sin}^2 45° + \text{cos}^2 45°}{\text{tan}^2 60°}$

(ii) $\dfrac{\text{sin 30° - sin 90° + 2 cos 0°}}{\text{tan 30° × tan 60°}}$

(iii) $\dfrac{4}{3} \text{tan}^2 30° + \text{sin}^2 60° - \text{3 cos}^2 60° + \dfrac{3}{4}\text{tan}^2 60° - 2\text{tan}^2 45°$.

Answer

(i) Solving,

$\Rightarrow \dfrac{\text{sin}^2 45° + \text{cos}^2 45°}{\text{tan}^2 60°} \\[1em] [\because \text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} = 1], \\[1em] \Rightarrow \dfrac{1}{(\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{1}{3}$

Hence, $\dfrac{\text{sin}^2 45° + \text{cos}^2 45°}{\text{tan}^2 60°} = \dfrac{1}{3}$.

(ii) Solving,

$\Rightarrow \dfrac{\text{sin 30° - sin 90° + 2 cos 0°}}{\text{tan 30° × tan 60°}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} - 1 + 2 \times 1}{\dfrac{1}{\sqrt{3}} \times \sqrt{3}} \\[1em] \Rightarrow \dfrac{\dfrac{1 - 2 + 4}{2}}{1} \\[1em] \Rightarrow \dfrac{3}{2}.$

Hence, $\dfrac{\text{sin 30° - sin 90° + 2 cos 0°}}{\text{tan 30° × tan 60°}} = \dfrac{3}{2}$.

(iii) Solving,

$\Rightarrow\dfrac{4}{3} \text{tan}^2 30° + \text{sin}^2 60° - \text{3 cos}^2 60° + \dfrac{3}{4}\text{tan}^2 60° - 2\text{tan}^2 45° \\[1em] \Rightarrow \dfrac{4}{3} \times \Big(\dfrac{1}{\sqrt{3}}\Big)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2 - 3 \times \Big(\dfrac{1}{2}\Big)^2 + \dfrac{3}{4} \times (\sqrt{3})^2 - 2 \times (1)^2 \\[1em] \Rightarrow \dfrac{4}{3} \times \dfrac{1}{3} + \dfrac{3}{4} - \dfrac{3}{4} + \dfrac{3}{4} \times 3 - 2 \\[1em] \Rightarrow \dfrac{4}{9} + \dfrac{9}{4} - 2 \\[1em] \Rightarrow \dfrac{16 + 81 - 72}{36} \\[1em] \Rightarrow \dfrac{25}{36}.$

Hence, $\dfrac{4}{3} \text{tan}^2 30° + \text{sin}^2 60° - \text{3 cos}^2 60° + \dfrac{3}{4}\text{tan}^2 60° - 2\text{tan}^2 45° = \dfrac{25}{36}$.

Find the values of

(i) $\dfrac{\text{sin 60°}}{\text{cos}^2 45°} - 3\text{ tan 30° + 5 cos 90°}$

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

(iii) $\dfrac{4}{5}\text{tan}^2 60° - \dfrac{2}{\text{sin}^2 30°} - \dfrac{3}{4}\text{tan}^2 30°$

Answer

(i) Solving,

$\Rightarrow \dfrac{\dfrac{\sqrt{3}}{2}}{\Big(\dfrac{1}{\sqrt{2}}\Big)^2} - 3 \times \dfrac{1}{\sqrt{3}} + 5 \times 0 \\[1em] \Rightarrow \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} - \sqrt{3} \\[1em] \Rightarrow \dfrac{2\sqrt{3}}{2} - \sqrt{3} \\[1em] \Rightarrow \sqrt{3} - \sqrt{3} \\[1em] \Rightarrow 0.$

Hence, $\dfrac{\text{sin 60°}}{\text{cos}^2 45°} - 3\text{ tan 30° + 5 cos 90°} = 0.$

(ii) Solving,

$\Rightarrow 2\sqrt{2}\text{ cos 45° cos 60°} + 2\sqrt{3} \text{ sin 30° tan 60° - cos 0°} \\[1em] \Rightarrow 2\sqrt{2} \times \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2} + 2 \times \sqrt{3} \times \dfrac{1}{2} \times \sqrt{3} - 1 \\[1em] \Rightarrow 1 + \sqrt{3} \times \sqrt{3} - 1 \\[1em] \Rightarrow 1 + 3 - 1 \\[1em] \Rightarrow 3.$

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

(iii) Solving,

$\Rightarrow \dfrac{4}{5}\text{tan}^2 60° - \dfrac{2}{\text{sin}^2 30°} - \dfrac{3}{4}\text{tan}^2 30° \\[1em] \Rightarrow \dfrac{4}{5} \times (\sqrt{3})^2 - \dfrac{2}{\Big(\dfrac{1}{2}\Big)^2} - \dfrac{3}{4} \times \Big(\dfrac{1}{\sqrt{3}}\Big)^2 \\[1em] \Rightarrow \dfrac{4}{5} \times 3 - 2 \times (2)^2 - \dfrac{3}{4} \times \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{12}{5} - 8 - \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{48 - 160 - 5}{20} \\[1em] \Rightarrow -\dfrac{117}{20}\\[1em] \Rightarrow -5\dfrac{17}{20}.$

Hence, $\dfrac{4}{5}\text{tan}^2 60° - \dfrac{2}{\text{sin}^2 30°} - \dfrac{3}{4}\text{tan}^2 30° = -5\dfrac{17}{20}$.

Prove that

(i) cos² 30° + sin 30° + tan² 45° = $2\dfrac{1}{4}$

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

(iii) cos 60° = cos² 30° - sin² 30°.

Answer

(i) Solving,

L.H.S. of the equation : cos² 30° + sin 30° + tan² 45° = $2\dfrac{1}{4}$.

$\Rightarrow \Big(\dfrac{\sqrt{3}}{2}\Big)^2 + \dfrac{1}{2} + (1)^2 \\[1em] \Rightarrow \dfrac{3}{4} + \dfrac{1}{2} + 1 \\[1em] \Rightarrow \dfrac{3 + 2 + 4}{4} \\[1em] \Rightarrow \dfrac{9}{4} \\[1em] \Rightarrow 2\dfrac{1}{4}.$

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

Hence, proved that cos² 30° + sin 30° + tan² 45° = $2\dfrac{1}{4}$.

(ii) Solving,

L.H.S. of the equation : 4(sin⁴ 30° + cos⁴ 60°) - 3(cos² 45° - sin² 90°) = 2

$\Rightarrow 4\Big[\Big(\dfrac{1}{2}\Big)^4 + \Big(\dfrac{1}{2}\Big)^4\Big] - 3\Big[\Big(\dfrac{1}{\sqrt{2}}\Big)^2 - (1)^2\Big] \\[1em] \Rightarrow 4\Big[\dfrac{1}{16} + \dfrac{1}{16}\Big] - 3\Big[\dfrac{1}{2} - 1\Big] \\[1em] \Rightarrow 4 \times \dfrac{2}{16} - 3 \times -\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{2} + \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{4}{2} \\[1em] \Rightarrow 2.$

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

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

(iii) Solving,

R.H.S. of the equation : cos 60° = cos² 30° - sin² 30°.

$\Rightarrow \Big(\dfrac{\sqrt{3}}{2}\Big)^2 - \Big(\dfrac{1}{2}\Big)^2 \\[1em] \Rightarrow \dfrac{3}{4} - \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{2}{4} \\[1em] \Rightarrow \dfrac{1}{2} \\[1em] \Rightarrow \text{cos 60°}.$

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

Hence, proved that cos 60° = cos² 30° - sin² 30°.

If x = 30°, verify that tan 2x = $\dfrac{\text{2 tan x}}{\text{1 - tan}^2 x}$.

Answer

To verify,

tan 2x = $\dfrac{\text{2 tan x}}{\text{1 - tan}^2 x}$.

Substituting value of x in L.H.S. of the above equation,

⇒ tan 2x = tan 2(30°) = tan 60° = $\sqrt{3}$.

Substituting value of x in R.H.S. of the equation.

$\Rightarrow \dfrac{\text{2 tan x}}{\text{1 - tan}^2 x} \\[1em] \Rightarrow \dfrac{2 \times \text{tan 30°}}{1 - \text{tan}^2 30°} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 - \Big(\dfrac{1}{\sqrt{3}}\Big)^2} \\[1em] \Rightarrow \dfrac{\dfrac{2}{\sqrt{3}}}{1 - \dfrac{1}{3}} \\[1em] \Rightarrow \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} \\[1em] \Rightarrow \dfrac{2 \times 3}{2 \times \sqrt{3}} \\[1em] \Rightarrow \sqrt{3}.$

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

Hence, proved that tan 2x = $\dfrac{\text{2 tan x}}{\text{1 - tan}^2 x}$.

If x = 15°, verify that 4 sin 2x cos 4x sin 6x = 1.

Answer

To verify,

4 sin 2x cos 4x sin 6x = 1

Substituting value of x in L.H.S. of the above equation.

⇒ 4 sin 2(15°) cos 4(15°) sin 6(15°)

⇒ 4 sin 30° cos 60° sin 90°

⇒ $4 \times \dfrac{1}{2} \times \dfrac{1}{2} \times 1$

⇒ 1.

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

Hence, proved that 4 sin 2x cos 4x sin 6x = 1.

Find the values of

(i) $\sqrt{\dfrac{1 - \text{cos}^2 30°}{1 - \text{sin}^2 30°}}$

(ii) $\dfrac{\text{sin 45° cos 45° cos 60°}}{\text{sin 60° cos 30° tan 45°}}$

Answer

(i) Solving,

$\Rightarrow \sqrt{\dfrac{1 - \text{cos}^2 30°}{1 - \text{sin}^2 30°}} \\[1em] \Rightarrow \sqrt{\dfrac{1 - \Big(\dfrac{\sqrt{3}}{2}\Big)^2}{1 - \Big(\dfrac{1}{2}\Big)^2}} \\[1em] \Rightarrow \sqrt{\dfrac{1 - \dfrac{3}{4}}{1 - \dfrac{1}{4}}} \\[1em] \Rightarrow \sqrt{\dfrac{\dfrac{1}{4}}{\dfrac{3}{4}}} \\[1em] \Rightarrow \sqrt{\dfrac{1 \times 4}{3 \times 4}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}}.$

Hence, $\sqrt{\dfrac{1 - \text{cos}^2 30°}{1 - \text{sin}^2 30°}} = \dfrac{1}{\sqrt{3}}$.

(ii) Solving,

$\Rightarrow \dfrac{\text{sin 45° cos 45° cos 60°}}{\text{sin 60° cos 30° tan 45°}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} \times 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{4}}{\dfrac{3}{4}} \\[1em] \Rightarrow \dfrac{1}{3}.$

Hence, $\dfrac{\text{sin 45° cos 45° cos 60°}}{\text{sin 60° cos 30° tan 45°}} = \dfrac{1}{3}.$

If θ = 30°, verify that

(i) sin 2θ = 2 sin θ cos θ

(ii) cos 2θ = 2 cos² θ - 1

(iii) sin 3θ = 3 sin θ - 4 sin³ θ

(iv) cos 3θ = 4 cos³ θ - 3 cos θ

Answer

(i) To verify,

sin 2θ = 2 sin θ cos θ

Substituting value of θ in L.H.S. of the equation we get :

⇒ sin 2θ = sin 2(30°) = sin 60° = $\dfrac{\sqrt{3}}{2}$.

Substituting value of θ in R.H.S. of the equation we get :

⇒ 2 sin θ cos θ = 2 sin 30° cos 30° = $2 \times \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}$.

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

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

(ii) To verify,

cos 2θ = 2 cos² θ - 1

Substituting value of θ in L.H.S. of the equation we get :

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

Substituting value of θ in R.H.S. of the equation we get :

⇒ 2cos² θ - 1 = 2cos² 30° - 1

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

= $2 \times \dfrac{3}{4} - 1 = \dfrac{3}{2} - 1 = \dfrac{1}{2}$.

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

Hence, proved that cos 2θ = 2cos² θ - 1.

(iii) To verify,

sin 3θ = 3 sin θ - 4 sin³ θ

Substituting value of θ in L.H.S. of the equation we get :

⇒ sin 3θ = sin 3(30°) = sin 90° = 1.

Substituting value of θ in R.H.S. of the equation we get :

⇒ 3 sin θ - 4 sin³ θ = 3 sin 30° - 4 sin³ 30°

= $3 \times \dfrac{1}{2} - 4 \times \Big(\dfrac{1}{2}\Big)^3$

= $\dfrac{3}{2} - \dfrac{4}{8}$

= $\dfrac{12 - 4}{8} = \dfrac{8}{8}$ = 1.

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

Hence, proved that sin 3θ = 3 sin θ - 4 sin³ θ.

(iv) Given,

Equation : cos 3θ = 4 cos³θ - 3 cos θ

Substituting θ = 30°, in L.H.S. of the given equation, we get :

⇒ cos 3θ = cos 3(30°) = cos 90° = 0.

Substituting θ = 30°, in R.H.S. of the given equation, we get :

⇒ 4 cos³θ - 3 cos θ = 4 cos³ 30° - 3 cos 30°

$= 4 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3 - 3 \times \dfrac{\sqrt{3}}{2}\\[1em] = 4 \times \dfrac{3\sqrt{3}}{8} - \dfrac{3\sqrt{3}}{2} \\[1em] = \dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2} \\[1em] = 0.$

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

Hence, proved that cos 3θ = 4 cos³θ - 3 cos θ.

If θ = 30°, find the ratio 2 sin θ : sin 2θ.

Answer

Substituting, θ = 30° in 2 sin θ : sin 2θ we get,

⇒ 2 sin 30° : sin 2(30°)

⇒ 2 sin 30° : sin 60°

⇒ $\dfrac{2 \times \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}$

⇒ $\dfrac{1}{\dfrac{\sqrt{3}}{2}}$

⇒ $\dfrac{2}{\sqrt{3}} = 2 : \sqrt{3}$.

Hence, 2 sin θ : sin 2θ = $\dfrac{2}{\sqrt{3}} = 2 : \sqrt{3}$.

By, means of an example, show that sin(A + B) ≠ sin A + sin B.

Answer

Let A = 30° and B = 60°.

Substituting values in sin(A + B) we get :

⇒ sin(A + B) = sin(30° + 60°) = sin 90° = 1.

Substituting values in sin A + sin B we get :

⇒ sin A + sin B = sin 30° + sin 60°

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

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

As, LHS ≠ RHS

Hence, proved that sin(A + B) ≠ sin A + sin B.

If A = 60° and B = 30°, verify that

(i) sin(A + B) = sin A cos B + cos A sin B

(ii) cos(A + B) = cos A cos B - sin A sin B

(iii) sin(A - B) = sin A cos B - cos A sin B

(iv) tan(A - B) = $\dfrac{\text{tan A - tan B}}{\text{1 + tan A tan B}}.$

Answer

(i) To verify,

sin(A + B) = sin A cos B + cos A sin B

Substituting values in L.H.S. of the above equation :

sin(A + B) = sin(60° + 30°) = sin 90° = 1.

Substituting values in R.H.S. of the equation :

sin A cos B + cos A sin B = sin 60° cos 30° + cos 60° sin 30°

$= \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{4}{4} = 1.$

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

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

(ii) To verify,

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

Substituting values in L.H.S. of equation :

cos(A + B) = cos(60° + 30°) = cos 90° = 0.

Substituting values in R.H.S. of equation :

cos A cos B - sin A sin B = cos 60° cos 30° - sin 60° sin 30°

$= \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4} \\[1em] = 0.$

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

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

(iii) To verify,

sin(A - B) = sin A cos B - cos A sin B

Substituting values in L.H.S. of equation :

sin(A - B) = sin(60° - 30°) = sin 30° = $\dfrac{1}{2}$.

Substituting values in R.H.S. of equation :

sin A cos B - cos A sin B = sin 60° cos 30° - cos 60° sin 30°

$= \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{3}{4} - \dfrac{1}{4} \\[1em] = \dfrac{2}{4} = \dfrac{1}{2}.$

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

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

(iv) To verify,

tan (A - B) = $\dfrac{\text{tan A - tan B}}{1 + \text{ tan A tan B}}$.

Substituting values in L.H.S. of equation :

tan (A - B) = tan (60° - 30°) = tan 30° = $\dfrac{1}{\sqrt{3}}$.

Substituting values in R.H.S. of equation :

$\dfrac{\text{tan A - tan B}}{1 + \text{ tan A tan B}} = \dfrac{\text{tan 60° - tan 30°}}{1 + \text{tan 60° tan 30°}} \\[1em] = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \dfrac{1}{\sqrt{3}}} \\[1em] = \dfrac{\dfrac{3 - 1}{\sqrt{3}}}{1 + 1} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{2} \\[1em] = \dfrac{1}{\sqrt{3}}.$

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

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

If 2θ is an acute angle and 2 sin 2θ = $\sqrt{3}$, find the value of θ.

Answer

Given,

⇒ 2 sin 2θ = $\sqrt{3}$

⇒ sin 2θ = $\dfrac{\sqrt{3}}{2}$

⇒ sin 2θ = sin 60°

⇒ 2θ = 60°

⇒ θ = 30°.

Hence, θ = 30°.

If 20° + x is an acute angle and cos(20° + x) = sin 60°, then find the value of x.

Answer

We know that,

sin 60° = $\dfrac{\sqrt{3}}{2}$ = cos 30°.

Given,

⇒ cos(20° + x) = sin 60°

⇒ cos(20° + x) = cos 30°

⇒ (20° + x) = 30°

⇒ x = 10°.

Hence, x = 10°.

If 3 sin² θ = $2\dfrac{1}{4}$ and θ is less than 90°, find the value of θ.

Answer

Given,

$\Rightarrow 3\text{sin}^2 \text{ θ} = \dfrac{9}{4} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{3}{4} \\[1em] \Rightarrow \text{sin} \text{ θ} = \pm \dfrac{\sqrt{3}}{2}.$

Since, θ is an acute angle.

∴ sin θ = $\dfrac{\sqrt{3}}{2}.$

⇒ sin θ = sin 60°

⇒ θ = 60°.

Hence, θ = 60°.

If θ is an acute angle and sin θ = cos θ, find the value of θ and hence, find the value of 2 tan² θ + sin² θ - 1.

Answer

Given,

sin θ = cos θ

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Substituting value in 2 tan² θ + sin² θ - 1 we get :

⇒ 2 tan² 45° + sin² 45° - 1

⇒ 2(1)² + $\Big(\dfrac{1}{\sqrt{2}}\Big)^2 -1$

⇒ 2 + $\dfrac{1}{2}$ - 1

⇒ $1\dfrac{1}{2}$.

Hence, 2 tan² θ + sin² θ - 1 = $1\dfrac{1}{2}$.

From the adjoining figure, find

(i) tan x°

(ii) x

(iii) cos x°

(iv) Without using Pythagoras theorem, find y.

Answer

(i) By formula,

tan x° = $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BC} = \sqrt{3}$.

Hence, tan x° = $\sqrt{3}$.

(ii) tan x° = $\sqrt{3}$

⇒ tan x° = tan 60°

⇒ x = 60.

Hence, x = 60.

(iii) Substituting value of x in cos x°, we get :

⇒ cos x° = cos 60°

⇒ cos x° = $\dfrac{1}{2}$.

Hence, cos x° = $\dfrac{1}{2}$.

(iv) Substituting value of x in sin x°, we get :

⇒ sin x° = sin 60° = $\dfrac{\sqrt{3}}{2}$.

By formula,

$\Rightarrow \text{sin x}\degree = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{y} \\[1em] \Rightarrow y = 2.$

Hence, y = 2.

If 3θ is an acute angle, solve the following equations for θ :

(i) 2 sin 3θ = $\sqrt{3}$

(ii) tan 3θ = 1.

Answer

(i) Given,

⇒ 2 sin 3θ = $\sqrt{3}$

⇒ sin 3θ = $\dfrac{\sqrt{3}}{2}$

⇒ sin 3θ = sin 60°

⇒ 3θ = 60°

⇒ θ = $\dfrac{60\degree}{3}$

⇒ θ = 20°.

Hence, θ = 20°.

(ii) Given,

⇒ tan 3θ = 1

⇒ tan 3θ = tan 45°

⇒ 3θ = 45°

⇒ θ = $\dfrac{45\degree}{3}$

⇒ θ = 15°.

Hence, θ = 15°.

If tan 3x = sin 45° cos 45° + sin 30°, find the value of x.

Answer

Given,

$\Rightarrow \text{tan 3x = sin 45° cos 45° + sin 30°} \\[1em] \Rightarrow \text{tan 3x} = \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} + \dfrac{1}{2} \\[1em] \Rightarrow \text{tan 3x} = \dfrac{1}{2} + \dfrac{1}{2} \\[1em] \Rightarrow \text{tan 3x} = 1 \\[1em] \Rightarrow \text{tan 3x = tan 45°} \\[1em] \Rightarrow 3x = 45° \\[1em] \Rightarrow x = \dfrac{45\degree}{3} \\[1em] \Rightarrow x = 15°.$

Hence, x = 15°.

If 4 cos² x° - 1 = 0 and 0 ≤ x ≤ 90, find

(i) x

(ii) sin² x° + cos² x°

(iii) cos² x° - sin² x°.

Answer

(i) Given,

$\Rightarrow \text{4 cos}^2 x° - 1 = 0 \\[1em] \Rightarrow \text{4 cos}^2 x° = 1 \\[1em] \Rightarrow \text{cos}^2 x° = \dfrac{1}{4} \\[1em] \Rightarrow \text{cos x°} = \sqrt{\dfrac{1}{4}} \\[1em] \Rightarrow \text{cos x°} = \pm \dfrac{1}{2}$

Since, x is an acute angle.

$\therefore \text{cos x°} = \dfrac{1}{2}$.

⇒ cos x° = cos 60°

⇒ x = 60.

Hence, x = 60.

(ii) Substituting value of x in sin² x° + cos² x° we get :

$\text{sin}^2 x° + \text{cos}^2 x° = \text{sin}^2 60° + \text{cos}^2 60° \\[1em] = \Big(\dfrac{\sqrt{3}}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{4}{4} \\[1em] = 1.$

Hence, sin² x° + cos² x° = 1.

(iii) Substituting value of x in cos² x° - sin² x° we get :

$\text{cos}^2 x° - \text{sin}^2 x° = \text{cos}^2 60° - \text{sin}^2 60° \\[1em] = \Big(\dfrac{1}{2}\Big)^2 - \Big(\dfrac{\sqrt{3}}{2}\Big)^2 \\[1em] = \dfrac{1}{4} - \dfrac{3}{4} \\[1em] = -\dfrac{2}{4} \\[1em] = -\dfrac{1}{2}.$

Hence, cos² x° - sin² x° = $-\dfrac{1}{2}$.

If sec θ = cosec θ and 0° ≤ θ ≤ 90°, find the value of θ.

Answer

Given,

$\Rightarrow \text{sec θ = cosec θ} \\[1em] \Rightarrow \dfrac{1}{\text{cos θ}} = \dfrac{1}{\text{sin θ}} \\[1em] \Rightarrow \dfrac{\text{sin θ}}{\text{cos θ}} = 1 \\[1em] \Rightarrow \text{tan θ} = 1 \\[1em] \Rightarrow \text{tan θ = tan 45°} \\[1em] \Rightarrow \text{θ} = 45°.$

Hence, θ = 45°.

If tan θ = cot θ and 0° ≤ θ ≤ 90°, find the value of θ.

Answer

Given,

As, θ is in the range of 0° ≤ θ ≤ 90°.

tan θ = cot θ [Only when θ = 45°].

Hence, θ = 45°.

If sin 3x = 1 and 0° ≤ 3x ≤ 90°, find the values of

(i) sin x

(ii) cos 2x

(iii) tan² x - sec² x.

Answer

Given,

⇒ sin 3x = 1

⇒ sin 3x = sin 90°

⇒ 3x = 90°

⇒ x = $\dfrac{90}{3}$

⇒ x = 30°.

(i) Substituting value of x in sin x, we get :

⇒ sin x = sin 30° = $\dfrac{1}{2}$.

Hence, sin x = $\dfrac{1}{2}$.

(ii) Substituting value of x in cos 2x, we get :

⇒ cos 2x = cos 60° = $\dfrac{1}{2}$.

Hence, cos x = $\dfrac{1}{2}$.

(iii) Substituting value of x in tan² x - sec² x, we get :

$\Rightarrow \text{tan}^2 x - \text{sec}^2 x = \text{tan}^2 60° - \text{sec}^2 60° \\[1em] = (\sqrt{3})^2 - (2)^2 \\[1em] = 3 - 4 \\[1em] = -1.$

Hence, tan² x - sec² x = -1.

If 3 tan² θ - 1 = 0, find cos 2θ, given that θ is acute.

Answer

Given,

⇒ 3 tan² θ - 1 = 0

⇒ tan² θ = $\dfrac{1}{3}$

⇒ tan θ = $\sqrt{\dfrac{1}{3}}$

⇒ tan θ = $\pm \dfrac{1}{\sqrt{3}}$

As, θ is acute.

∴ tan θ = $\dfrac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

cos 2θ = cos 60° = $\dfrac{1}{2}$

Hence, cos 2θ = $\dfrac{1}{2}$.

If sin x + cos y = 1, x = 30° and y is acute angle, find the value of y.

Answer

Given,

⇒ sin x + cos y = 1

⇒ sin 30° + cos y = 1

⇒ $\dfrac{1}{2}$ + cos y = 1

⇒ cos y = 1 - $\dfrac{1}{2}$

⇒ cos y = $\dfrac{1}{2}$

⇒ cos y = cos 60°.

Hence, y = 60°.

If sin(A + B) = $\dfrac{\sqrt{3}}{2}$ = cos(A - B), 0° < A + B ≤ 90° (A > B), find the values of A and B.

Answer

Given,

⇒ sin(A + B) = $\dfrac{\sqrt{3}}{2}$

⇒ sin(A + B) = sin 60°

⇒ A + B = 60° ..........(1)

Also,

⇒ cos(A - B) = $\dfrac{\sqrt{3}}{2}$

⇒ cos(A - B) = cos 30°

⇒ A - B = 30° ..........(2)

Adding, (1) and (2), we get :

⇒ (A + B) + (A - B) = 60° + 30°

⇒ A + A + B - B = 90°

⇒ 2A = 90°

⇒ A = 45°.

Substituting value of A in (1), we get :

⇒ A + B = 60°

⇒ 45° + B = 60°

⇒ B = 60° - 45°

⇒ B = 15°.

Hence, A = 45° and B = 15°.

If the length of each side of a rhombus is 8 cm and its one angle is 60°, then find the lengths of the diagonals of the rhombus.

Answer

We know that the diagonals of a rhombus bisect the opposite angles and are perpendicular to each other.

∴ ∠OAB = $\dfrac{60°}{2}$ = 30°.

In right ∠AOB,

⇒ sin 30° = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{OB}{AB}$

⇒ $\dfrac{1}{2} = \dfrac{OB}{AB}$

⇒ OB = $\dfrac{AB}{2}$

⇒ OB = $\dfrac{8}{2}$ = 4 cm.

As diagonals of rhombus bisect each other.

∴ BD = 2 OB = 2 × 4 = 8 cm.

cos 30° = $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{OA}{AB}$

⇒ $\dfrac{\sqrt{3}}{2} = \dfrac{OA}{AB}$

⇒ OA = $\dfrac{AB\sqrt{3}}{2}$

⇒ OA = $\dfrac{8\sqrt{3}}{2} = 4\sqrt{3}$.

As diagonals of rhombus bisect each other.

∴ AC = 2 OA = $2 \times 4\sqrt{3} = 8\sqrt{3}$.

Hence, the length of the diagonals of the rhombus are 8 cm and $8\sqrt{3}$ cm.

In the right-angled triangle ABC, ∠C = 90° and ∠B = 60°. If AC = 6 cm, find the lengths of the sides BC and AB.

Answer

From figure,

$\text{sin 60°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AC}{AB} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{6}{AB} \\[1em] \Rightarrow AB = \dfrac{12}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{12}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{12\sqrt{3}}{3} \\[1em] \Rightarrow AB = 4\sqrt{3} \text{ cm}. \\[1em] \text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{AC}{BC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{6}{BC} \\[1em] \Rightarrow BC = \dfrac{6}{\sqrt{3}} = 2\sqrt{3}.$

Hence, AB = $4\sqrt{3}$ cm and BC = $2\sqrt{3}$ cm.

In the adjoining figure, AP is a man of height 1.8 m and BQ is a building 13.8 m high. If the man sees the top of the building by focussing his binoculars at an angle of 30° to the horizontal, find the distance of the man from the building.

Answer

Let AB = d meters, then PC = d meters.

From right-angled △PCQ,

tan 30° = $\dfrac{CQ}{PC}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{BQ - BC}{d}$

From figure,

BC = AP = 1.8 m

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{13.8 - 1.8}{d}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{12}{d}$

⇒ d = $12\sqrt{3}$ meters.

Hence, distance of man from building is $12\sqrt{3}$ meters.

In the adjoining figure, ABC is a triangle in which ∠B = 45° and ∠C = 60°. If AD ⊥ BC and BC = 8m, find the length of the altitude AD.

Answer

In △ABD,

⇒ tan 45° = $\dfrac{AD}{BD}$

⇒ 1 = $\dfrac{AD}{BD}$

⇒ BD = AD.

In △ADC,

⇒ tan 60° = $\dfrac{AD}{DC}$

⇒ $\sqrt{3} = \dfrac{AD}{DC}$

⇒ DC = $\dfrac{AD}{\sqrt{3}}$

From figure,

BC = BD + DC

$\Rightarrow 8 = AD + \dfrac{AD}{\sqrt{3}} \\[1em] \Rightarrow 8 = \dfrac{\sqrt{3}AD + AD}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{AD(\sqrt{3} + 1)}{\sqrt{3}} = 8 \\[1em] \Rightarrow AD = \dfrac{8\sqrt{3}}{\sqrt{3} + 1}$

Multiplying numerator and denominator by $(\sqrt{3} - 1)$

$\Rightarrow AD = \dfrac{8\sqrt{3}}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} \\[1em] \Rightarrow AD = \dfrac{8(3 - \sqrt{3})}{(\sqrt{3})^2 - (1)^2} \space [\because a^2 - b^2 = (a+b)(a-b)] \\[1em] \Rightarrow AD = \dfrac{8(3 - \sqrt{3})}{3 - 1} \\[1em] \Rightarrow AD = \dfrac{8(3 - \sqrt{3})}{2}\\[1em] \Rightarrow AD = 4(3 - \sqrt{3}) \text{ m}$

Hence, AD = $4(3 - \sqrt{3})$ m.