SEBA Class 10 Advanced Maths Chapter 7.2 Solutions: Trigonometry

Looking for reliable and step-by-step SEBA Class 10 Advanced Maths Chapter 7.2 Solutions? In this article, we provide complete, easy-to-understand solutions to all textual questions of Exercise 7.2 (Trigonometry) from the SEBA Class 10 New Advanced Mathematics textbook. This article helps you get full marks from this chapter in your upcoming HSLC board examination.

See More:
SEBA Class 10 Advanced Maths Chapter 7.1
SEBA Class 10 Advanced Maths Chapter 7.3

1. Using the formulae for compound angles show that:

(i) $\sin(45^\circ – \alpha) = \cos(45^\circ + \alpha) = \frac{1}{\sqrt{2}}(\cos\alpha – \sin\alpha)$

Soln :

$\sin(45^\circ – \alpha) = \sin 45^\circ \cos\alpha – \cos 45^\circ \sin\alpha$

$= \frac{1}{\sqrt{2}}\cos\alpha – \frac{1}{\sqrt{2}}\sin\alpha$

$= \frac{1}{\sqrt{2}}(\cos\alpha – \sin\alpha)$ — (i)

Again,

$\cos(45^\circ + \alpha) = \cos 45^\circ \cos\alpha – \sin 45^\circ \sin\alpha$

$= \frac{1}{\sqrt{2}}\cos\alpha – \frac{1}{\sqrt{2}}\sin\alpha$

$= \frac{1}{\sqrt{2}}(\cos\alpha – \sin\alpha)$ — (ii)

From (i) and (ii), we get:

$\sin(45^\circ – \alpha) = \cos(45^\circ + \alpha) = \frac{1}{\sqrt{2}}(\cos\alpha – \sin\alpha)$

Showed.

(ii) $\sin(45^\circ + \alpha) + \cos(45^\circ – \alpha) = \sqrt{2}(\cos\alpha + \sin\alpha)$

Soln :

$\text{L.H.S.} = \sin(45^\circ + \alpha) + \cos(45^\circ – \alpha)$

$= (\sin 45^\circ \cos\alpha + \cos 45^\circ \sin\alpha) + (\cos 45^\circ \cos\alpha + \sin 45^\circ \sin\alpha)$

$= \frac{1}{\sqrt{2}}\cos\alpha + \frac{1}{\sqrt{2}}\sin\alpha + \frac{1}{\sqrt{2}}\cos\alpha + \frac{1}{\sqrt{2}}\sin\alpha$

$= \frac{2}{\sqrt{2}}\cos\alpha + \frac{2}{\sqrt{2}}\sin\alpha$

$= \sqrt{2}\cos\alpha + \sqrt{2}\sin\alpha$

$= \sqrt{2}(\cos\alpha + \sin\alpha)$

$= \text{R.H.S.}$

Showed.

(iii) $\sin(\theta – 60^\circ) + \cos(\theta – 30^\circ) = \sin\theta$

Soln :

$\text{L.H.S.} = \sin(\theta – 60^\circ) + \cos(\theta – 30^\circ)$

$= (\sin\theta \cos 60^\circ – \cos\theta \sin 60^\circ) + (\cos\theta \cos 30^\circ + \sin\theta \sin 30^\circ)$

$= \sin\theta \cdot \frac{1}{2} – \cos\theta \cdot \frac{\sqrt{3}}{2} + \cos\theta \cdot \frac{\sqrt{3}}{2} + \sin\theta \cdot \frac{1}{2}$

$= \frac{1}{2}\sin\theta + \frac{1}{2}\sin\theta$

$= \sin\theta$

$= \text{R.H.S.}$

Showed.

(iv) $\cos 70^\circ \cos 10^\circ + \sin 70^\circ \sin 10^\circ = \frac{1}{2}$

Soln :

$\text{L.H.S.} = \cos 70^\circ \cos 10^\circ + \sin 70^\circ \sin 10^\circ$

$= \cos(70^\circ – 10^\circ)$

$= \cos 60^\circ$

$= \frac{1}{2}$

$= \text{R.H.S.}$

Showed.

(v) $\sin 45^\circ \cos 15^\circ – \cos 45^\circ \sin 15^\circ = \frac{1}{2}$

Soln :

$\text{L.H.S.} = \sin 45^\circ \cos 15^\circ – \cos 45^\circ \sin 15^\circ$

$= \sin(45^\circ – 15^\circ)$

$= \sin 30^\circ$

$= \frac{1}{2}$

$= \text{R.H.S.}$

Showed.

(vi) $\tan 10^\circ + \tan 35^\circ + \tan 10^\circ \tan 35^\circ = 1$

Soln :

We know that:$\tan 45^\circ = 1$

$\implies \tan(10^\circ + 35^\circ) = 1$

$\implies \frac{\tan 10^\circ + \tan 35^\circ}{1 – \tan 10^\circ \tan 35^\circ} = 1$

$\implies \tan 10^\circ + \tan 35^\circ = 1 \times (1 – \tan 10^\circ \tan 35^\circ)$

$\implies \tan 10^\circ + \tan 35^\circ = 1 – \tan 10^\circ \tan 35^\circ$

$\implies \tan 10^\circ + \tan 35^\circ + \tan 10^\circ \tan 35^\circ = 1$

Showed.

(vii) $\sin(\alpha + \beta)\cos(\alpha + \gamma) – \cos(\alpha + \beta)\sin(\alpha + \gamma) = \sin(\beta – \gamma)$

Soln :

$\text{L.H.S.} = \sin(\alpha + \beta)\cos(\alpha + \gamma) – \cos(\alpha + \beta)\sin(\alpha + \gamma)$

$= \sin[(\alpha + \beta) – (\alpha + \gamma)]$ [Let $A = \alpha + \beta$ and $B = \alpha + \gamma$]

$= \sin(\alpha + \beta – \alpha – \gamma)$

$= \sin(\beta – \gamma)$

$= \text{R.H.S.}$

Showed.

(viii) $\tan(45^\circ + \alpha)\tan(45^\circ – \alpha) = 1$

Soln :

$\text{L.H.S.} = \tan(45^\circ + \alpha)\tan(45^\circ – \alpha)$

$= \left(\frac{\tan 45^\circ + \tan\alpha}{1 – \tan 45^\circ \tan\alpha}\right) \left(\frac{\tan 45^\circ – \tan\alpha}{1 + \tan 45^\circ \tan\alpha}\right)$

$= \left(\frac{1 + \tan\alpha}{1 – \tan\alpha}\right) \left(\frac{1 – \tan\alpha}{1 + \tan\alpha}\right) \quad [\because \tan 45^\circ = 1]$

$= 1$

$= \text{R.H.S.}$

Showed.

2. If $\tan\theta – \tan\phi = x$ and $\cot\phi – \cot\theta = y$, prove that:

$\cot(\theta – \phi) = \frac{1}{x} + \frac{1}{y}$

Soln :

Here,$\tan\theta – \tan\phi = x$

$\implies \frac{1}{\cot\theta} – \frac{1}{\cot\phi} = x$

$\implies \frac{\cot\phi – \cot\theta}{\cot\phi \cdot \cot\theta} = x$

$\implies \frac{y}{\cot\phi \cdot \cot\theta} = x \quad [\because \cot\phi – \cot\theta = y]$

$\implies \frac{y}{x} = \cot\phi \cdot \cot\theta$

Now,$\text{L.H.S.} = \cot(\theta – \phi)$

$= \frac{\cot\theta \cdot \cot\phi + 1}{\cot\phi – \cot\theta}$

$= \frac{\frac{y}{x} + 1}{y}$

$= \frac{\frac{y + x}{x}}{y}$$= \frac{y + x}{xy}$

$= \frac{y}{xy} + \frac{x}{xy}$$= \frac{1}{x} + \frac{1}{y}$

$= \text{R.H.S.}$

$\therefore \cot(\theta – \phi) = \frac{1}{x} + \frac{1}{y}$

Proved.

3. If $\theta$ and $\phi$ are acute positive angles and $\sin\theta = \frac{3}{5}$, $\sin\phi = \frac{5}{13}$, then find the value of $\sin(\theta + \phi)$, $\cos(\theta – \phi)$ and $\tan(\theta + \phi)$.

Soln :

Here,$\sin\theta = \frac{3}{5}, \quad \sin\phi = \frac{5}{13}$

Since $\theta$ and $\phi$ are positive acute angles: $\cos\theta = \sqrt{1 – \sin^2\theta}$ $= \sqrt{1 – \frac{9}{25}}$ $= \sqrt{\frac{16}{25}}$ $= \frac{4}{5}$

$\cos\phi = \sqrt{1 – \sin^2\phi}$

$= \sqrt{1 – \frac{25}{169}}$

$= \sqrt{\frac{144}{169}}$

$= \frac{12}{13}$

(i) Now,

$\sin(\theta + \phi) = \sin\theta \cos\phi + \cos\theta \sin\phi$

$= \frac{3}{5} \times \frac{12}{13} + \frac{4}{5} \times \frac{5}{13}$

$= \frac{36}{65} + \frac{20}{65}$

$= \frac{56}{65}$

(ii) $\cos(\theta – \phi) = \cos\theta \cos\phi + \sin\theta \sin\phi$

$= \frac{4}{5} \times \frac{12}{13} + \frac{3}{5} \times \frac{5}{13}$

$= \frac{48}{65} + \frac{15}{65}$

$= \frac{63}{65}$

(iii) We know, $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

$\tan\phi = \frac{\sin\phi}{\cos\phi} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$

Now,$\tan(\theta + \phi) = \frac{\tan\theta + \tan\phi}{1 – \tan\theta \tan\phi}$

$= \frac{\frac{3}{4} + \frac{5}{12}}{1 – \frac{3}{4} \times \frac{5}{12}}$

$= \frac{\frac{9 + 5}{12}}{\frac{48 – 15}{48}}$

$= \frac{14}{12} \times \frac{48}{33}$

$= \frac{56}{33}$

$\therefore \sin(\theta + \phi) = \frac{56}{65}, \quad \cos(\theta – \phi) = \frac{63}{65}, \quad \tan(\theta + \phi) = \frac{56}{33}$

4. If $\theta, \phi$ are positive acute angles and $\sin\theta = \frac{3}{5}$, $\cos\phi = \frac{9}{41}$, then show that: $\frac{\cos(\theta + \phi)}{\sin(\theta – \phi)} = \frac{12}{19}$

Soln :

Given,$\sin\theta = \frac{3}{5}, \quad \cos\phi = \frac{9}{41}$

We know, $\cos\theta = \sqrt{1 – \sin^2\theta}$

$= \sqrt{1 – \frac{9}{25}}$

$= \sqrt{\frac{16}{25}}$

$= \frac{4}{5}$

Again, $\sin\phi = \sqrt{1 – \cos^2\phi}$

$= \sqrt{1 – \left(\frac{9}{41}\right)^2}$

$= \sqrt{1 – \frac{81}{1681}}$

$= \sqrt{\frac{1600}{1681}}$

$= \frac{40}{41}$

Now,

$\text{L.H.S.} = \frac{\cos(\theta + \phi)}{\sin(\theta – \phi)}$

$= \frac{\cos\theta \cos\phi – \sin\theta \sin\phi}{\sin\theta \cos\phi – \cos\theta \sin\phi}$

$= \frac{\frac{4}{5} \times \frac{9}{41} – \frac{3}{5} \times \frac{40}{41}}{\frac{3}{5} \times \frac{9}{41} – \frac{4}{5} \times \frac{40}{41}}$

$= \frac{\frac{36}{205} – \frac{120}{205}}{\frac{27}{205} – \frac{160}{205}}$

$= \frac{-\frac{84}{205}}{-\frac{133}{205}}$

$= \frac{-84}{-133}$$= \frac{84}{133}$

$= \frac{12}{19}$

$= \text{R.H.S.}$

Showed.

5. Show that:

$\frac{\sin(A – B)}{\cos A \cos B} + \frac{\sin(B – C)}{\cos B \cos C} + \frac{\sin(C – A)}{\cos C \cos A} = 0$

Soln :

$\text{L.H.S.} = \frac{\sin(A – B)}{\cos A \cos B} + \frac{\sin(B – C)}{\cos B \cos C} + \frac{\sin(C – A)}{\cos C \cos A}$

$= \frac{\sin A \cos B – \cos A \sin B}{\cos A \cos B} + \frac{\sin B \cos C – \cos B \sin C}{\cos B \cos C} + \frac{\sin C \cos A – \cos C \sin A}{\cos C \cos A}$

$= \left(\frac{\sin A \cos B}{\cos A \cos B} – \frac{\cos A \sin B}{\cos A \cos B}\right) + \left(\frac{\sin B \cos C}{\cos B \cos C} – \frac{\cos B \sin C}{\cos B \cos C}\right) + \left(\frac{\sin C \cos A}{\cos C \cos A} – \frac{\cos C \sin A}{\cos C \cos A}\right)$

$= (\tan A – \tan B) + (\tan B – \tan C) + (\tan C – \tan A)$

$= \tan A – \tan B + \tan B – \tan C + \tan C – \tan A$$= 0$

$= \text{R.H.S.}$

Showed.

6. Evaluate:

(i) $\sin\theta + \sin(\theta + 120^\circ) + \sin(\theta – 120^\circ)$

Soln :

$\sin\theta + \sin(\theta + 120^\circ) + \sin(\theta – 120^\circ)$

$= \sin\theta + \sin(\theta + 120^\circ) + \sin[-(120^\circ – \theta)]$

$= \sin\theta + \sin(\theta + 120^\circ) – \sin(120^\circ – \theta) \quad [\because \sin(-\theta) = -\sin\theta]$

$= \sin\theta + \sin[90^\circ + (\theta + 30^\circ)] – \sin[90^\circ – (\theta – 30^\circ)]$

$= \sin\theta + \cos(\theta + 30^\circ) – \cos(\theta – 30^\circ)$

$= \sin\theta + (\cos\theta \cos 30^\circ – \sin\theta \sin 30^\circ) – (\cos\theta \cos 30^\circ + \sin\theta \sin 30^\circ)$

$= \sin\theta + \cos\theta \cos 30^\circ – \sin\theta \sin 30^\circ – \cos\theta \cos 30^\circ – \sin\theta \sin 30^\circ$

$= \sin\theta – 2\sin\theta \sin 30^\circ$

$= \sin\theta – 2\sin\theta \times \frac{1}{2}$

$= \sin\theta – \sin\theta$

$= 0$

(ii) $\cos\theta + \cos(\theta + 120^\circ) + \cos(120^\circ – \theta)$

Soln :

$\cos\theta + \cos(\theta + 120^\circ) + \cos(120^\circ – \theta)$

$= \cos\theta + \cos[90^\circ + (\theta + 30^\circ)] + \cos[90^\circ – (\theta – 30^\circ)]$

$= \cos\theta – \sin(\theta + 30^\circ) + \sin(\theta – 30^\circ) \quad [\because \cos(90^\circ + A) = -\sin A, \cos(90^\circ – A) = \sin A]$

$= \cos\theta + \sin(\theta – 30^\circ) – \sin(\theta + 30^\circ)$

$= \cos\theta + (\sin\theta \cos 30^\circ – \cos\theta \sin 30^\circ) – (\sin\theta \cos 30^\circ + \cos\theta \sin 30^\circ)$

$= \cos\theta + \sin\theta \cos 30^\circ – \cos\theta \sin 30^\circ – \sin\theta \cos 30^\circ – \cos\theta \sin 30^\circ$$= \cos\theta – 2\cos\theta \sin 30^\circ$

$= \cos\theta – 2\cos\theta \times \frac{1}{2}$

$= \cos\theta – \cos\theta$

$= 0$

(iii) $\sin 135^\circ \cos 145^\circ + \cos 135^\circ \sin 145^\circ$

Soln :

$= \sin 135^\circ \cos 145^\circ + \cos 135^\circ \sin 145^\circ$

$= \sin(135^\circ + 145^\circ)$

$= \sin 280^\circ$

$= \sin(3 \times 90^\circ + 10^\circ)$

$= -\cos 10^\circ$

7. Show that:

(i) $\sin^2 A + \sin^2(A – B) – 2\sin A \cos B \sin(A – B) = \sin^2 B$

Soln :

$\text{L.H.S.} = \sin^2 A + \sin^2(A – B) – 2\sin A \cos B \sin(A – B)$

$= \sin^2 A + \sin(A – B)[\sin(A – B) – 2\sin A \cos B]$

$= \sin^2 A + \sin(A – B)[\sin A \cos B – \cos A \sin B – 2\sin A \cos B]$

$= \sin^2 A + \sin(A – B)[-\sin A \cos B – \cos A \sin B]$

$= \sin^2 A – \sin(A – B)(\sin A \cos B + \cos A \sin B)$

$= \sin^2 A – \sin(A – B)\sin(A + B) \quad [\because \sin(A + B) = \sin A \cos B + \cos A \sin B]$

$= \sin^2 A – (\sin^2 A – \sin^2 B) \quad [\because \sin(A + B)\sin(A – B) = \sin^2 A – \sin^2 B]$

$= \sin^2 A – \sin^2 A + \sin^2 B$

$= \sin^2 B$

$= \text{R.H.S.}$

Showed.

(ii) $\cos^2 B + \cos^2(A – B) – 2\cos(A – B)\cos A \cos B = \sin^2 A$

Soln :

$\text{L.H.S.} = \cos^2 B + \cos^2(A – B) – 2\cos(A – B)\cos A \cos B$

$= \cos^2 B + \cos(A – B)[\cos(A – B) – 2\cos A \cos B]$

$= \cos^2 B + \cos(A – B)[\cos A \cos B + \sin A \sin B – 2\cos A \cos B]$

$= \cos^2 B + \cos(A – B)[-\cos A \cos B + \sin A \sin B]$

$= \cos^2 B – \cos(A – B)[\cos A \cos B – \sin A \sin B]$

$= \cos^2 B – \cos(A – B)\cos(A + B) \quad [\because \cos(A + B) = \cos A \cos B – \sin A \sin B]$

$= \cos^2 B – (\cos^2 B – \sin^2 A) \quad [\because \cos(A + B)\cos(A – B) = \cos^2 B – \sin^2 A]$

$= \cos^2 B – \cos^2 B + \sin^2 A$

$= \sin^2 A$

$= \text{R.H.S.}$

Showed.

8. Show that:

$(\cos\alpha – \sin\alpha)(\cos 2\alpha – \sin 2\alpha) = \cos\alpha – \sin 3\alpha$

Soln :

$\text{L.H.S.} = (\cos\alpha – \sin\alpha)(\cos 2\alpha – \sin 2\alpha)$

$= \cos\alpha \cos 2\alpha – \cos\alpha \sin 2\alpha – \sin\alpha \cos 2\alpha + \sin\alpha \sin 2\alpha$

$= (\cos 2\alpha \cos\alpha + \sin 2\alpha \sin\alpha) – (\sin 2\alpha \cos\alpha + \cos 2\alpha \sin\alpha)$

$= \cos(2\alpha – \alpha) – \sin(2\alpha + \alpha)$

$= \cos\alpha – \sin 3\alpha$

$= \text{R.H.S.}$

Showed.

9. Show that:

$1 + \tan 2A \tan A = \tan 2A \cot A – 1 = \sec 2A$

Soln :

Part 1:$\text{L.H.S.} = 1 + \tan 2A \tan A$

$= 1 + \frac{\sin 2A}{\cos 2A} \cdot \frac{\sin A}{\cos A}$

$= \frac{\cos 2A \cos A + \sin 2A \sin A}{\cos 2A \cos A}$

$= \frac{\cos(2A – A)}{\cos 2A \cos A} \quad [\because \cos(A – B) = \cos A \cos B + \sin A \sin B]$

$= \frac{\cos A}{\cos 2A \cos A}$

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

$= \sec 2A$ — (i)

Part 2:$\text{Middle Term} = \tan 2A \cot A – 1$

$= \frac{\sin 2A}{\cos 2A} \cdot \frac{\cos A}{\sin A} – 1$

$= \frac{\sin 2A \cos A – \cos 2A \sin A}{\cos 2A \sin A}$

$= \frac{\sin(2A – A)}{\cos 2A \sin A} \quad [\because \sin(A – B) = \sin A \cos B – \cos A \sin B]$

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

$= \frac{1}{\cos 2A}$ $= \sec 2A$ — (ii)

From (i) and (ii), we get:

$1 + \tan 2A \tan A = \tan 2A \cot A – 1 = \sec 2A$

Showed.

10. Show that:(i) $\frac{\cos 9^\circ + \sin 9^\circ}{\cos 9^\circ – \sin 9^\circ} = \tan 54^\circ$

Soln :

$\text{L.H.S.} = \frac{\cos 9^\circ + \sin 9^\circ}{\cos 9^\circ – \sin 9^\circ}$

Dividing the numerator and denominator by $\cos 9^\circ$:

$= \frac{\frac{\cos 9^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 9^\circ}}{\frac{\cos 9^\circ}{\cos 9^\circ} – \frac{\sin 9^\circ}{\cos 9^\circ}}$

$= \frac{1 + \tan 9^\circ}{1 – \tan 9^\circ}$

$= \frac{\tan 45^\circ + \tan 9^\circ}{1 – \tan 45^\circ \tan 9^\circ} \quad [\because \tan 45^\circ = 1]$

$= \tan(45^\circ + 9^\circ) \quad \left[\because \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B}\right]$

$= \tan 54^\circ$

$= \text{R.H.S.}$

Showed.

(ii) $\frac{\cos 10^\circ – \sin 10^\circ}{\cos 10^\circ + \sin 10^\circ} = \tan 35^\circ$

Soln :

$\text{L.H.S.} = \frac{\cos 10^\circ – \sin 10^\circ}{\cos 10^\circ + \sin 10^\circ}$

Dividing the numerator and denominator by $\cos 10^\circ$:

$= \frac{\frac{\cos 10^\circ}{\cos 10^\circ} – \frac{\sin 10^\circ}{\cos 10^\circ}}{\frac{\cos 10^\circ}{\cos 10^\circ} + \frac{\sin 10^\circ}{\cos 10^\circ}}$

$= \frac{1 – \tan 10^\circ}{1 + \tan 10^\circ}$

$= \frac{\tan 45^\circ – \tan 10^\circ}{1 + \tan 45^\circ \tan 10^\circ} \quad [\because \tan 45^\circ = 1]$

$= \tan(45^\circ – 10^\circ) \quad \left[\because \tan(A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}\right]$

$= \tan 35^\circ$

$= \text{R.H.S.}$

Showed.

11. If $A + B + C = 180^\circ$, then prove that: $\tan A + \tan B + \tan C = \tan A \tan B \tan C$

Soln :

Given,$A + B + C = 180^\circ$

$\implies A + B = 180^\circ – C$

Taking tangent on both sides, we get

$\implies \tan(A + B) = \tan(180^\circ – C)$

$\implies \frac{\tan A + \tan B}{1 – \tan A \tan B} = -\tan C$

$\implies \tan A + \tan B = -\tan C (1 – \tan A \tan B)$

$\implies \tan A + \tan B = -\tan C + \tan A \tan B \tan C$

$\implies \tan A + \tan B + \tan C = \tan A \tan B \tan C$

Proved.

12. If $A + B + C = 180^\circ$ and $\sin A + \sin B \sin C = 0$, then prove that:$2\tan B + \tan C = 0$  

Soln :  

Given,$A + B + C = 180^\circ$

$\implies A = 180^\circ – (B + C)$

$\implies \sin A = \sin[180^\circ – (B + C)]$ [Taking sine on both sides]

$\implies \sin A = \sin(B + C) \quad [\because \sin(180^\circ – \theta) = \sin\theta]$

$\implies \sin A = \sin B \cos C + \cos B \sin C$ — (i)

Also given,

$\sin A + \sin B \cos C = 0$

$\implies \sin A = -\sin B \cos C$

Substituting the value of $\sin A$ in equation (i), we get

$\implies -\sin B \cos C = \sin B \cos C + \cos B \sin C$

$\implies -2\sin B \cos C = \cos B \sin C$

$\implies 2\sin B \cos C + \cos B \sin C = 0$

$\implies \frac{2\sin B \cos C}{\cos B \cos C} + \frac{\cos B \sin C}{\cos B \cos C} = 0$ [Dividing both sides by $\cos B \cos C$]

$\implies 2\tan B + \tan C = 0$

Proved.

13. If $\alpha + \beta = \theta$, $\alpha – \beta = x$, and $\tan\alpha : \tan\beta = k$, then show that $\sin x = \frac{k – 1}{k + 1} \cdot \sin\theta$

Soln :

Given, $\frac{\tan\alpha}{\tan\beta} = k$

$\implies \frac{\frac{\sin\alpha}{\cos\alpha}}{\frac{\sin\beta}{\cos\beta}} = k$

$\implies \frac{\sin\alpha \cos\beta}{\cos\alpha \sin\beta} = k$

Now, subtracting $1$ from both sides, we gwt

$\frac{\sin\alpha \cos\beta}{\cos\alpha \sin\beta} – 1 = k – 1$

$\implies \frac{\sin\alpha \cos\beta – \cos\alpha \sin\beta}{\cos\alpha \sin\beta} = k – 1$

$\implies \frac{\sin(\alpha – \beta)}{\cos\alpha \sin\beta} = k – 1 \longrightarrow \text{(i)}$

And, adding $1$ to both sides, we gwt

$\frac{\sin\alpha \cos\beta}{\cos\alpha \sin\beta} + 1 = k + 1$

$\implies \frac{\sin\alpha \cos\beta + \cos\alpha \sin\beta}{\cos\alpha \sin\beta} = k + 1$

$\implies \frac{\sin(\alpha + \beta)}{\cos\alpha \sin\beta} = k + 1 \longrightarrow \text{(ii)}$

Now, dividing equation $\text{(i)}$ by $\text{(ii)}$ [$\text{(i)} \div \text{(ii)}$], we get

$\frac{k – 1}{k + 1} = \frac{\frac{\sin(\alpha – \beta)}{\cos\alpha \sin\beta}}{\frac{\sin(\alpha + \beta)}{\cos\alpha \sin\beta}}$

$\implies \frac{k – 1}{k + 1} = \frac{\sin(\alpha – \beta)}{\sin(\alpha + \beta)}$

$\implies \frac{k – 1}{k + 1} = \frac{\sin x}{\sin\theta} \quad [\because \alpha – \beta = x, \, \alpha + \beta = \theta]$

$\therefore \sin x = \frac{k – 1}{k + 1} \cdot \sin\theta$

Showed.

14. If $\tan\theta = \frac{a\sin\alpha + b\sin\beta}{a\cos\alpha + b\cos\beta}$, then prove that: $a\sin(\theta – \alpha) + b\sin(\theta – \beta) = 0$

Soln :

Given,

$\tan\theta = \frac{a\sin\alpha + b\sin\beta}{a\cos\alpha + b\cos\beta}$

$\implies \frac{\sin\theta}{\cos\theta} = \frac{a\sin\alpha + b\sin\beta}{a\cos\alpha + b\cos\beta}$

$\implies \sin\theta(a\cos\alpha + b\cos\beta) = \cos\theta(a\sin\alpha + b\sin\beta)$ [By cross-multiplication]

$\implies a\sin\theta \cos\alpha + b\sin\theta \cos\beta = a\cos\theta \sin\alpha + b\cos\theta \sin\beta$

$\implies a\sin\theta \cos\alpha – a\cos\theta \sin\alpha + b\sin\theta \cos\beta – b\cos\theta \sin\beta = 0$

$\implies a(\sin\theta \cos\alpha – \cos\theta \sin\alpha) + b(\sin\theta \cos\beta – \cos\theta \sin\beta) = 0$

$\implies a\sin(\theta – \alpha) + b\sin(\theta – \beta) = 0$

Proved.

15. If $m\sin(\theta + \alpha) = n\sin(\theta + \beta)$, then show that: $\cot\theta = \frac{m\cos\alpha – n\cos\beta}{n\sin\beta – m\sin\alpha}$

Soln :

Given,

$m\sin(\theta + \alpha) = n\sin(\theta + \beta)$

$\implies m(\sin\theta \cos\alpha + \cos\theta \sin\alpha) = n(\sin\theta \cos\beta + \cos\theta \sin\beta)$

$\implies m\sin\theta \cos\alpha + m\cos\theta \sin\alpha = n\sin\theta \cos\beta + n\cos\theta \sin\beta$

$\implies m\sin\theta \cos\alpha – n\sin\theta \cos\beta = n\cos\theta \sin\beta – m\cos\theta \sin\alpha$

$\implies \sin\theta(m\cos\alpha – n\cos\beta) = \cos\theta(n\sin\beta – m\sin\alpha)$

$\implies \frac{m\cos\alpha – n\cos\beta}{n\sin\beta – m\sin\alpha} = \frac{\cos\theta}{\sin\theta}$

$\implies \frac{m\cos\alpha – n\cos\beta}{n\sin\beta – m\sin\alpha} = \cot\theta$

$\implies \cot\theta = \frac{m\cos\alpha – n\cos\beta}{n\sin\beta – m\sin\alpha}$

Showed.

16. If $\tan\theta = \frac{x\sin\phi}{1 – x\cos\phi}$ and $\tan\phi = \frac{y\sin\theta}{1 – y\cos\theta}$, then show that:$\frac{x}{y} = \frac{\sin\theta}{\sin\phi}$

Soln :

From the first relation:

$\tan\theta = \frac{x\sin\phi}{1 – x\cos\phi}$

$\implies \tan\theta(1 – x\cos\phi) = x\sin\phi$

$\implies \tan\theta – x\tan\theta \cos\phi = x\sin\phi$

$\implies \tan\theta = x\sin\phi + x\tan\theta \cos\phi$

$\implies \tan\theta = x(\sin\phi + \tan\theta \cos\phi)$

$\implies x = \frac{\tan\theta}{\sin\phi + \tan\theta \cos\phi}$ — (i)

From the second relation:

$\tan\phi = \frac{y\sin\theta}{1 – y\cos\theta}$

$\implies \tan\phi(1 – y\cos\theta) = y\sin\theta$

$\implies \tan\phi – y\tan\phi \cos\theta = y\sin\theta$

$\implies \tan\phi = y(\sin\theta + \tan\phi \cos\theta)$

$\implies y = \frac{\tan\phi}{\sin\theta + \tan\phi \cos\theta}$ — (ii)

Now,

$\text{L.H.S.} = \frac{x}{y}$

$= \frac{\frac{\tan\theta}{\sin\phi + \tan\theta \cos\phi}}{\frac{\tan\phi}{\sin\theta + \tan\phi \cos\theta}}$

$= \frac{\tan\theta(\sin\theta + \tan\phi \cos\theta)}{\tan\phi(\sin\phi + \tan\theta \cos\phi)}$

$= \frac{\frac{\sin\theta}{\cos\theta}\left(\sin\theta + \frac{\sin\phi}{\cos\phi}\cos\theta\right)}{\frac{\sin\phi}{\cos\phi}\left(\sin\phi + \frac{\sin\theta}{\cos\theta}\cos\phi\right)}$

$= \frac{\frac{\sin\theta}{\cos\theta}\left(\frac{\sin\theta \cos\phi + \cos\theta \sin\phi}{\cos\phi}\right)}{\frac{\sin\phi}{\cos\phi}\left(\frac{\sin\phi \cos\theta + \cos\phi \sin\theta}{\cos\theta}\right)}$

$= \frac{\frac{\sin\theta}{\cos\theta \cos\phi}\sin(\theta + \phi)}{\frac{\sin\phi}{\cos\theta \cos\phi}\sin(\theta + \phi)}$

$= \frac{\sin\theta}{\sin\phi}$

$= \text{R.H.S.}$

$\therefore \frac{x}{y} = \frac{\sin\theta}{\sin\phi}$

Proved.

17. Express as sum or difference of trigonometrical ratios:

(i) $2\sin 54^\circ \cos 30^\circ$

Soln :

We know that, $2\sin A \cos B = \sin(A + B) + \sin(A – B)$:

$2\sin 54^\circ \cos 30^\circ$

$= \sin(54^\circ + 30^\circ) + \sin(54^\circ – 30^\circ)$

$= \sin 84^\circ + \sin 24^\circ$

(ii) $2\cos(135^\circ + \theta)\cos(135^\circ – \theta)$

Soln :

We know that, $2\cos A \cos B = \cos(A + B) + \cos(A – B)$:

$2\cos(135^\circ + \theta)\cos(135^\circ – \theta)$

$= \cos[(135^\circ + \theta) + (135^\circ – \theta)] + \cos[(135^\circ + \theta) – (135^\circ – \theta)]$

$= \cos 270^\circ + \cos 2\theta$

(iii) $2\sin 5\theta \sin 7\theta$

Soln :

We know that, $2\sin A \sin B = \cos(A – B) – \cos(A + B)$:

$2\sin 5\theta \sin 7\theta$

$= \cos(5\theta – 7\theta) – \cos(5\theta + 7\theta)$

$= \cos(-2\theta) – \cos 12\theta \quad [\because \cos(-\theta) = \cos\theta]$

$= \cos 2\theta – \cos 12\theta$

(iv) $\sin 80^\circ \sin 40^\circ$

Soln :

$\sin 80^\circ \sin 40^\circ$

$= \frac{1}{2} [2\sin 80^\circ \sin 40^\circ]$

We know that, $2\sin A \sin B = \cos(A – B) – \cos(A + B)$:

$= \frac{1}{2}[\cos(80^\circ – 40^\circ) – \cos(80^\circ + 40^\circ)]$

$= \frac{1}{2}[\cos 40^\circ – \cos 120^\circ]$

18. Express as product of trigonometrical ratios:

(i) $\sin 75^\circ + \sin 15^\circ$

Soln :

Using the relation $\sin C + \sin D = 2\sin\frac{C + D}{2}\cos\frac{C – D}{2}$:

$\sin 75^\circ + \sin 15^\circ$

$= 2\sin\frac{75^\circ + 15^\circ}{2}\cos\frac{75^\circ – 15^\circ}{2}$

$= 2\sin 45^\circ \cos 30^\circ$

(ii) $\sin 6\theta – \sin 4\theta$

Soln :

Using the relation $\sin C – \sin D = 2\cos\frac{C + D}{2}\sin\frac{C – D}{2}$:

$\sin 6\theta – \sin 4\theta$

$= 2\cos\frac{6\theta + 4\theta}{2}\sin\frac{6\theta – 4\theta}{2}$

$= 2\cos 5\theta \sin\theta$

(iii) $\cos 3\theta – \cos 7\theta$

Soln :

Using the relation $\cos C – \cos D = 2\sin\frac{C + D}{2}\sin\frac{D – C}{2}$:

$\cos 3\theta – \cos 7\theta$

$= 2\sin\frac{3\theta + 7\theta}{2}\sin\frac{7\theta – 3\theta}{2}$

$= 2\sin 5\theta \sin 2\theta$

(iv) $\sin 105^\circ + \cos 15^\circ$

Soln :

$\sin 105^\circ + \cos 15^\circ$

$= \sin(90^\circ + 15^\circ) + \cos 15^\circ$

$= \cos 15^\circ + \cos 15^\circ \quad [\because \sin(90^\circ + \theta) = \cos\theta]$

$= 2\cos 15^\circ$

19. Show that:

(i) $\frac{\sin A + \sin B}{\cos A + \cos B} = \tan\frac{A + B}{2}$

Soln :

$\text{L.H.S.} = \frac{\sin A + \sin B}{\cos A + \cos B}$

$= \frac{2\sin\frac{A + B}{2}\cos\frac{A – B}{2}}{2\cos\frac{A + B}{2}\cos\frac{A – B}{2}}$$= \frac{\sin\frac{A + B}{2}}{\cos\frac{A + B}{2}}$

$= \tan\frac{A + B}{2}$ $= \text{R.H.S.}$

Showed.

(ii) $\frac{\cos 4\theta – \cos 6\theta}{\sin 4\theta + \sin 6\theta} = \tan\theta$

Soln :

$\text{L.H.S.} = \frac{\cos 4\theta – \cos 6\theta}{\sin 4\theta + \sin 6\theta}$

$= \frac{2\sin\frac{4\theta + 6\theta}{2}\sin\frac{6\theta – 4\theta}{2}}{2\sin\frac{4\theta + 6\theta}{2}\cos\frac{4\theta – 6\theta}{2}}$

$= \frac{2\sin 5\theta \sin\theta}{2\sin 5\theta \cos(-\theta)}$

$= \frac{\sin\theta}{\cos\theta} \quad [\because \cos(-\theta) = \cos\theta]$

$= \tan\theta$$= \text{R.H.S.}$

Showed.

20. Evaluate:

(i) $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$

Soln :

$\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$

$= 2\cos\frac{100^\circ + 20^\circ}{2}\cos\frac{100^\circ – 20^\circ}{2} + \cos 140^\circ$

$= 2\cos 60^\circ \cos 40^\circ + \cos 140^\circ$

$= 2\left(\frac{1}{2}\right)\cos 40^\circ + \cos 140^\circ \quad \left[\because \cos 60^\circ = \frac{1}{2}\right]$

$= \cos 40^\circ + \cos 140^\circ$

$= 2\cos\frac{140^\circ + 40^\circ}{2}\cos\frac{140^\circ – 40^\circ}{2}$

$= 2\cos 90^\circ \cos 50^\circ$

$= 2 \times 0 \times \cos 50^\circ \quad [\because \cos 90^\circ = 0]$

$= 0$

(ii) $\sin 20^\circ + \sin 40^\circ – \cos 10^\circ$

Soln :

$\sin 20^\circ + \sin 40^\circ – \cos 10^\circ$

$= 2\sin\frac{40^\circ + 20^\circ}{2}\cos\frac{40^\circ – 20^\circ}{2} – \cos 10^\circ$

$= 2\sin 30^\circ \cos 10^\circ – \cos 10^\circ$

$= 2\left(\frac{1}{2}\right)\cos 10^\circ – \cos 10^\circ \quad \left[\because \sin 30^\circ = \frac{1}{2}\right]$

$= \cos 10^\circ – \cos 10^\circ$

$= 0$

(iii) $\sin 80^\circ \cos 20^\circ + \sin 45^\circ \cos 145^\circ + \sin 55^\circ \cos 245^\circ$

Soln :

$\sin 80^\circ \cos 20^\circ + \sin 45^\circ \cos 145^\circ + \sin 55^\circ \cos 245^\circ$

$= \sin 80^\circ \cos 20^\circ + \sin 45^\circ \cos(90^\circ + 55^\circ) + \sin 55^\circ \cos(180^\circ + 65^\circ)$

$= \sin 80^\circ \cos 20^\circ – \sin 45^\circ \sin 55^\circ – \sin 55^\circ \cos 65^\circ $

$= \frac{1}{2}[2\sin 80^\circ \cos 20^\circ – 2\sin 55^\circ \sin 45^\circ – 2\cos 65^\circ \sin 55^\circ]$

$= \frac{1}{2}[\{\sin(80^\circ + 20^\circ) + \sin(80^\circ – 20^\circ)\} – \{\cos(55^\circ – 45^\circ) – \cos(55^\circ + 45^\circ)\} – \{\sin(65^\circ + 55^\circ) – \sin(65^\circ – 55^\circ)\}]$

$= \frac{1}{2}[\sin 100^\circ + \sin 60^\circ – \cos 10^\circ + \cos 100^\circ – \sin 120^\circ + \sin 10^\circ]$

$= \frac{1}{2}[\sin(90^\circ + 10^\circ) + \frac{\sqrt{3}}{2} – \cos 10^\circ + \cos(90^\circ + 10^\circ) – \sin(180^\circ – 60^\circ) + \sin 10^\circ]$

$= \frac{1}{2}[\cos 10^\circ + \frac{\sqrt{3}}{2} – \cos 10^\circ – \sin 10^\circ – \sin 60^\circ + \sin 10^\circ]$

$= \frac{1}{2}\left[\frac{\sqrt{3}}{2} – \frac{\sqrt{3}}{2}\right]$

$= 0$

(iv) $\cos 40^\circ \cos 100^\circ \cos 160^\circ$

Soln :

$\cos 40^\circ \cos 100^\circ \cos 160^\circ$

$= \frac{1}{2}[2\cos 100^\circ \cos 40^\circ] \cos 160^\circ$

$= \frac{1}{2}[\cos(100^\circ + 40^\circ) + \cos(100^\circ – 40^\circ)] \cos 160^\circ$

$= \frac{1}{2}[\cos 140^\circ + \cos 60^\circ] \cos 160^\circ$

$= \frac{1}{2}\left[\cos 140^\circ + \frac{1}{2}\right] \cos 160^\circ $

$= \frac{1}{2}\cos 140^\circ \cos 160^\circ + \frac{1}{4}\cos 160^\circ$

$= \frac{1}{4}[2\cos 160^\circ \cos 140^\circ] + \frac{1}{4}\cos 160^\circ$

$= \frac{1}{4}[\cos(160^\circ + 140^\circ) + \cos(160^\circ – 140^\circ)] + \frac{1}{4}\cos 160^\circ$

$= \frac{1}{4}[\cos 300^\circ + \cos 20^\circ] + \frac{1}{4}\cos 160^\circ$

$= \frac{1}{4}\cos(360^\circ – 60^\circ) + \frac{1}{4}\cos 20^\circ + \frac{1}{4}\cos(180^\circ – 20^\circ)$

$= \frac{1}{4}\cos 60^\circ + \frac{1}{4}\cos 20^\circ – \frac{1}{4}\cos 20^\circ \quad [\because \cos(360^\circ – \theta) = \cos\theta, \, \cos(180^\circ – \theta) = -\cos\theta]$

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

$= \frac{1}{8}$

21. Show that: $\cos 95^\circ + \cos 25^\circ = \frac{1}{\sqrt{2}}(\cos 10^\circ + \sin 10^\circ)$

Soln :

$\text{L.H.S.} = \cos 95^\circ + \cos 25^\circ$

$= 2\cos\frac{95^\circ + 25^\circ}{2}\cos\frac{95^\circ – 25^\circ}{2}$

$= 2\cos 60^\circ \cos 35^\circ$

$= 2 \times \frac{1}{2} \times \cos 35^\circ$

$= \cos 35^\circ$

$= \cos(45^\circ – 10^\circ)$

$= \cos 45^\circ \cos 10^\circ + \sin 45^\circ \sin 10^\circ $

$= \frac{1}{\sqrt{2}}\cos 10^\circ + \frac{1}{\sqrt{2}}\sin 10^\circ$

$= \frac{1}{\sqrt{2}}(\cos 10^\circ + \sin 10^\circ)$$= \text{R.H.S.}$

Showed.