SEBA Class 10 Advanced Maths Chapter 7.3 Solutions: Trigonometry

Get free SEBA Class 10 Advanced Maths Chapter 7.3 Solutions for the upcoming HSLC exam. In this article, we solved all the textual questions of Exercise 7.3 (Multiple Angles) from the SEBA Class 10 New Advanced Mathematics book. We provide 100% reliable and step-by-step solutions for all questions in a very simple way.

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SEBA Class 10 Advanced Maths Chapter 7.1
SEBA Class 10 Advanced Maths Chapter 7.2

1.(i) If $\sin\theta = \frac{1}{3}$ and $0^\circ < \theta < 90^\circ$, then find the value of $\sin 2\theta$, $\cos 2\theta$ and $\tan 2\theta$.

Soln :

Given, $\sin\theta = \frac{1}{3}$

Since $0^\circ < \theta < 90^\circ$, $\cos\theta$ is positive:

∴ $\cos\theta = \sqrt{1 – \sin^2\theta}$

$= \sqrt{1 – \left(\frac{1}{3}\right)^2}$

$= \sqrt{1 – \frac{1}{9}}$

$= \sqrt{\frac{8}{9}}$

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

Now, we know

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

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

$= \frac{4\sqrt{2}}{9}$

Again, we know

$\cos 2\theta = \cos^2\theta – \sin^2\theta$

$= \left(\frac{2\sqrt{2}}{3}\right)^2 – \left(\frac{1}{3}\right)^2$

$= \frac{8}{9} – \frac{1}{9}$

$= \frac{7}{9}$

Also, we know

$\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta}$

$= \frac{\frac{4\sqrt{2}}{9}}{\frac{7}{9}}$

$= \frac{4\sqrt{2}}{7}$

$\therefore \sin 2\theta = \frac{4\sqrt{2}}{9}, \quad \cos 2\theta = \frac{7}{9}, \quad \tan 2\theta = \frac{4\sqrt{2}}{7}$

1. (ii) If $\sin 2\theta = \frac{12}{13}$, then find $\tan\theta$.

Soln :

We know that,

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

Given, $\sin 2\theta = \frac{12}{13}$

$\implies \frac{2\tan\theta}{1 + \tan^2\theta} = \frac{12}{13}$

$\implies 12(1 + \tan^2\theta) = 26\tan\theta$

$\implies 12 + 12\tan^2\theta = 26\tan\theta$

$\implies 12\tan^2\theta – 26\tan\theta + 12 = 0$

$\implies 2(6\tan^2\theta – 13\tan\theta + 6) = 0$

$\implies 6\tan^2\theta – 13\tan\theta + 6 = 0$

$\implies 6\tan^2\theta – (9 + 4)\tan\theta + 6 = 0$

$\implies 6\tan^2\theta – 9\tan\theta – 4\tan\theta + 6 = 0$

$\implies 3\tan\theta(2\tan\theta – 3) – 2(2\tan\theta – 3) = 0$

$\implies (2\tan\theta – 3)(3\tan\theta – 2) = 0$

Either,

$2\tan\theta – 3 = 0 \implies \tan\theta = \frac{3}{2}$

Or,

$3\tan\theta – 2 = 0 \implies \tan\theta = \frac{2}{3}$

$\therefore \tan\theta = \frac{3}{2} \text{ or } \frac{2}{3}$

2. Show that:

(i) $\tan\theta = \frac{\sin 2\theta}{1 + \cos 2\theta}$

Soln :

We know that, $\sin 2\theta = 2\sin\theta\cos\theta$ and $1 + \cos 2\theta = 2\cos^2\theta$

$\text{R.H.S.} = \frac{\sin 2\theta}{1 + \cos 2\theta}$

$= \frac{2\sin\theta\cos\theta}{2\cos^2\theta}$

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

$= \tan\theta$

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

$\therefore \tan\theta = \frac{\sin 2\theta}{1 + \cos 2\theta}$

Showed.

(ii) $\frac{1 – \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta} = \tan\theta$

Soln :

$\text{L.H.S.} = \frac{(1 – \cos 2\theta) + \sin 2\theta}{(1 + \cos 2\theta) + \sin 2\theta}$

We know that $1 – \cos 2\theta = 2\sin^2\theta$, $1 + \cos 2\theta = 2\cos^2\theta$, and $\sin 2\theta = 2\sin\theta\cos\theta$:

$= \frac{2\sin^2\theta + 2\sin\theta\cos\theta}{2\cos^2\theta + 2\sin\theta\cos\theta}$

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

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

$= \tan\theta$

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

Showed.

(iii) $\frac{\sin A\cos A + \sin B\cos B}{\sin^2 A – \sin^2 B} = \cot(A – B)$

Soln :

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

$= \frac{2\sin A\cos A + 2\sin B\cos B}{2(\sin^2 A – \sin^2 B)}$ [Multiplying numerator and denominator by $2$]

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

$= \frac{2\sin\left(\frac{2A + 2B}{2}\right)\cos\left(\frac{2A – 2B}{2}\right)}{2\sin(A + B)\sin(A – B)}$ [Since,  $\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$]

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

$= \frac{\cos(A – B)}{\sin(A – B)}$

$= \cot(A – B)$

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

Showed.

(iv) $\frac{1 + \sin 2A}{\cos 2A} = \frac{\cos A + \sin A}{\cos A – \sin A} = \tan(45^\circ + A)$

Soln :

• First part:

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

  $= \frac{\cos^2 A + \sin^2 A + 2\sin A\cos A}{\cos^2 A – \sin^2 A} \quad [\because 1 = \cos^2 A + \sin^2 A]$

  $= \frac{(\cos A + \sin A)^2}{(\cos A + \sin A)(\cos A – \sin A)}$

  $= \frac{\cos A + \sin A}{\cos A – \sin A}$ — (i)

• Second part:

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

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

  $= \frac{1 + \tan A}{1 – \tan A}$

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

  $= \tan(45^\circ + A)$ — (ii)

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

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

Showed.

(v) $\cos 5\theta = 16\cos^5\theta – 20\cos^3\theta + 5\cos\theta$

(Note: Follows the method of textbook Example 6)

Soln :

$\text{L.H.S.} = \cos 5\theta$

$= \cos(3\theta + 2\theta)$

$= \cos 3\theta\cos 2\theta – \sin 3\theta\sin 2\theta$

Using the multiple formulas like $\cos 3\theta = 4\cos^3\theta – 3\cos\theta$, $\sin 3\theta = 3\sin\theta – 4\sin^3\theta$, $\cos 2\theta = 2\cos^2\theta – 1$, and $\sin 2\theta = 2\sin\theta\cos\theta$:

$= (4\cos^3\theta – 3\cos\theta)(2\cos^2\theta – 1) – (3\sin\theta – 4\sin^3\theta)(2\sin\theta\cos\theta)$

$= (8\cos^5\theta – 4\cos^3\theta – 6\cos^3\theta + 3\cos\theta) – 2\sin^2\theta\cos\theta(3 – 4\sin^2\theta)$

$= 8\cos^5\theta – 10\cos^3\theta + 3\cos\theta – 2(1 – \cos^2\theta)\cos\theta[3 – 4(1 – \cos^2\theta)]$

$= 8\cos^5\theta – 10\cos^3\theta + 3\cos\theta – 2(\cos\theta – \cos^3\theta)(4\cos^2\theta – 1)$

$= 8\cos^5\theta – 10\cos^3\theta + 3\cos\theta – 2(4\cos^3\theta – \cos\theta – 4\cos^5\theta + \cos^3\theta)$

$= 8\cos^5\theta – 10\cos^3\theta + 3\cos\theta – 2(-4\cos^5\theta + 5\cos^3\theta – \cos\theta)$

$= 8\cos^5\theta – 10\cos^3\theta + 3\cos\theta + 8\cos^5\theta – 10\cos^3\theta + 2\cos\theta$

$= 16\cos^5\theta – 20\cos^3\theta + 5\cos\theta$

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

Showed.

(vi) $\cos^2\alpha + \cos^2(\alpha + 120^\circ) + \cos^2(\alpha + 240^\circ) = \frac{3}{2}$

Soln :

$\text{L.H.S.} = \cos^2\alpha + \cos^2(\alpha + 120^\circ) + \cos^2(\alpha + 240^\circ)$

$= \cos^2\alpha + [\cos\{180^\circ – (60^\circ – \alpha)\}]^2 + [\cos\{180^\circ + (60^\circ + \alpha)\}]^2$

$= \cos^2\alpha + [-\cos(60^\circ – \alpha)]^2 + [-\cos(60^\circ + \alpha)]^2$

$= \cos^2\alpha + (\cos 60^\circ\cos\alpha + \sin 60^\circ\sin\alpha)^2 + (\cos 60^\circ\cos\alpha – \sin 60^\circ\sin\alpha)^2$

$= \cos^2\alpha + 2[\cos^2 60^\circ\cos^2\alpha + \sin^2 60^\circ\sin^2\alpha]$

$= \cos^2\alpha + 2\left[\left(\frac{1}{2}\right)^2\cos^2\alpha + \left(\frac{\sqrt{3}}{2}\right)^2\sin^2\alpha\right]$

$= \cos^2\alpha + 2\left[\frac{1}{4}\cos^2\alpha + \frac{3}{4}\sin^2\alpha\right]$

$= \cos^2\alpha + \frac{1}{2}\cos^2\alpha + \frac{3}{2}\sin^2\alpha$

$= \frac{3}{2}\cos^2\alpha + \frac{3}{2}\sin^2\alpha$

$= \frac{3}{2}(\cos^2\alpha + \sin^2\alpha)$

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

$= \frac{3}{2}$

Showed.

(vii) $\frac{1}{\sin 10^\circ} – \frac{\sqrt{3}}{\cos 10^\circ} = 4$

(Note: Follows the method of textbook Example 2)

Soln :

$\text{L.H.S.} = \frac{1}{\sin 10^\circ} – \frac{\sqrt{3}}{\cos 10^\circ}$

$= \frac{\cos 10^\circ – \sqrt{3}\sin 10^\circ}{\sin 10^\circ\cos 10^\circ}$

$= 2 \times \frac{\frac{1}{2}\cos 10^\circ – \frac{\sqrt{3}}{2}\sin 10^\circ}{\sin 10^\circ\cos 10^\circ}$

$= 2 \times \frac{\sin 30^\circ\cos 10^\circ – \cos 30^\circ\sin 10^\circ}{\sin 10^\circ\cos 10^\circ} \quad \left[\because \sin 30^\circ = \frac{1}{2}, \, \cos 30^\circ = \frac{\sqrt{3}}{2}\right]$

$= 2 \times \frac{\sin(30^\circ – 10^\circ)}{\sin 10^\circ\cos 10^\circ}$

$= 2 \times \frac{\sin 20^\circ}{\sin 10^\circ\cos 10^\circ}$

$= 2 \times \frac{2\sin 10^\circ\cos 10^\circ}{\sin 10^\circ\cos 10^\circ}$ [Since, $\sin 20^\circ = 2\sin 10^\circ\cos 10^\circ$]

$= 4$

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

Showed.

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

Soln :

$\text{L.H.S.} = \frac{1}{\sin 15^\circ} – \frac{1}{\cos 15^\circ}$

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

$= 2 \times \frac{\frac{1}{\sqrt{2}}\cos 15^\circ – \frac{1}{\sqrt{2}}\sin 15^\circ}{2\sin 15^\circ\cos 15^\circ} \times \sqrt{2}$

$= 2\sqrt{2} \times \frac{\sin 45^\circ\cos 15^\circ – \cos 45^\circ\sin 15^\circ}{\sin 30^\circ} \quad [\because 2\sin 15^\circ\cos 15^\circ = \sin 30^\circ]$

$= 2\sqrt{2} \times \frac{\sin(45^\circ – 15^\circ)}{\sin 30^\circ}$

$= 2\sqrt{2} \times \frac{\sin 30^\circ}{\sin 30^\circ}$

$= 2\sqrt{2}$

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

Showed.