SEBA Class 10 Maths Chapter 8.2 Introduction to Trigonometry (New Book)
Get the Free SEBA Class 10 Maths Chapter 8.2 Introduction to Trigonometry (New Book) Solution. This article provides complete solutions to Exercise 8.2 in a simple way based on the new SEBA Class 10 Maths textbook. These Class 10 Maths Chapter 8.2 solutions will help you understand the basics of Introduction to Trigonometry and prepare for the upcoming HSLC examination.
See More: Chapter 8.1 Introduction to Trigonometry
1. Evaluate the following:
(i) $\sin 60^{\circ} \cos 30^{\circ} + \sin 30^{\circ} \cos 60^{\circ}$
Solution:
$\sin 60^{\circ} \cos 30^{\circ} + \sin 30^{\circ} \cos 60^{\circ}$
$= \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right)$
$= \frac{3}{4} + \frac{1}{4}$
$= \frac{3 + 1}{4}$
$= \frac{4}{4}$
$= 1$
(ii) Evaluate: $2\tan^{2} 45^{\circ} + \cos^{2} 30^{\circ} – \sin^{2} 60^{\circ}$
Solution:
$2\tan^{2} 45^{\circ} + \cos^{2} 30^{\circ} – \sin^{2} 60^{\circ}$
$= 2 \times (1)^{2} + \left(\frac{\sqrt{3}}{2}\right)^{2} – \left(\frac{\sqrt{3}}{2}\right)^{2}$
$= 2 + \frac{3}{4} – \frac{3}{4}$
$= 2$
(iii) Evaluate: $\frac{\cos 45^{\circ}}{\sec 30^{\circ} + cosec 30^{\circ}}$
Solution:
$\frac{\cos 45^{\circ}}{\sec 30^{\circ} + cosec 30^{\circ}}$
$= \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$
$= \frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}}$
$= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 + 2\sqrt{3}}$
$= \frac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}}$
$= \frac{\sqrt{3} \times (2\sqrt{2} – 2\sqrt{6})}{(2\sqrt{2} + 2\sqrt{6}) \times (2\sqrt{2} – 2\sqrt{6})}$
$= \frac{2\sqrt{6} – 2\sqrt{18}}{(2\sqrt{2})^{2} – (2\sqrt{6})^{2}}$
$= \frac{2\sqrt{6} – 6\sqrt{2}}{8 – 24}$
$= \frac{2(\sqrt{6} – 3\sqrt{2})}{-16}$
$= \frac{\sqrt{6} – 3\sqrt{2}}{-8}$
$= \frac{3\sqrt{2} – \sqrt{6}}{8}$
(iv) Evaluate: $\frac{\sin 30^{\circ} + \tan 45^{\circ} – cosec 60^{\circ}}{\sec 30^{\circ} + \cos 60^{\circ} + \cot 45^{\circ}}$
$\frac{\sin 30^{\circ} + \tan 45^{\circ} – cosec 60^{\circ}}{\sec 30^{\circ} + \cos 60^{\circ} + \cot 45^{\circ}}$
$= \frac{\frac{1}{2} + 1 – \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1}$
$= \frac{\frac{3}{2} – \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{3}{2}}$
$= \frac{\frac{3\sqrt{3} – 4}{2\sqrt{3}}}{\frac{4 + 3\sqrt{3}}{2\sqrt{3}}}$
$= \frac{3\sqrt{3} – 4}{2\sqrt{3}} \times \frac{2\sqrt{3}}{4 + 3\sqrt{3}}$
$= \frac{3\sqrt{3} – 4}{3\sqrt{3} + 4}$
$= \frac{(3\sqrt{3} – 4) \times (3\sqrt{3} – 4)}{(3\sqrt{3} + 4) \times (3\sqrt{3} – 4)}$
$= \frac{(3\sqrt{3} – 4)^{2}}{(3\sqrt{3})^{2} – (4)^{2}}$
$= \frac{27 – 24\sqrt{3} + 16}{27 – 16}$
$= \frac{43 – 24\sqrt{3}}{11}$
(v) Evaluate: $\frac{5\cos^{2} 60^{\circ} + 4\sec^{2} 30^{\circ} – \tan^{2} 45^{\circ}}{\sin^{2} 30^{\circ} + \cos^{2} 30^{\circ}}$
$\frac{5\cos^{2} 60^{\circ} + 4\sec^{2} 30^{\circ} – \tan^{2} 45^{\circ}}{\sin^{2} 30^{\circ} + \cos^{2} 30^{\circ}}$
$= \frac{5 \times \left(\frac{1}{2}\right)^{2} + 4 \times \left(\frac{2}{\sqrt{3}}\right)^{2} – (1)^{2}}{\left(\frac{1}{2}\right)^{2} + \left(\frac{\sqrt{3}}{2}\right)^{2}}$
$= \frac{\frac{5}{4} + \frac{16}{3} – 1}{\frac{1}{4} + \frac{3}{4}}$
$= \frac{\frac{15 + 64 – 12}{12}}{\frac{4}{4}}$
$= \frac{\frac{67}{12}}{1}$
$= \frac{67}{12}$
(vi) Evaluate: $\frac{cosec 30^{\circ} + cosec 60^{\circ} + cosec 90^{\circ}}{\sec 0^{\circ} + \sec 30^{\circ} + \sec 60^{\circ}}$
$\frac{cosec 30^{\circ} + cosec 60^{\circ} + cosec 90^{\circ}}{\sec 0^{\circ} + \sec 30^{\circ} + \sec 60^{\circ}}$
$= \frac{2 + \frac{2}{\sqrt{3}} + 1}{1 + \frac{2}{\sqrt{3}} + 2}$
$= \frac{3 + \frac{2}{\sqrt{3}}}{3 + \frac{2}{\sqrt{3}}}$
$= 1$
(vii) Evaluate: $\frac{4\sin 30^{\circ} + \cos^{2} 45^{\circ} – 4\cos^{2} 60^{\circ}}{2\sin 30^{\circ}\cos 30^{\circ} + \tan^{2} 45^{\circ}}$
$\frac{4\sin 30^{\circ} + \cos^{2} 45^{\circ} – 4\cos^{2} 60^{\circ}}{2\sin 30^{\circ}\cos 30^{\circ} + \tan^{2} 45^{\circ}}$
$= \frac{4 \times \left(\frac{1}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)^{2} – 4 \times \left(\frac{1}{2}\right)^{2}}{2 \times \left(\frac{1}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right) + (1)^{2}}$
$= \frac{2 + \frac{1}{2} – 1}{\frac{\sqrt{3}}{2} + 1}$
$= \frac{1 + \frac{1}{2}}{\frac{\sqrt{3} + 2}{2}}$
$= \frac{\frac{3}{2}}{\frac{\sqrt{3} + 2}{2}}$
$= \frac{3}{2} \times \frac{2}{\sqrt{3} + 2}$
$= \frac{3}{2 + \sqrt{3}}$
$= \frac{3 \times (2 – \sqrt{3})}{(2 + \sqrt{3}) \times (2 – \sqrt{3})}$
$= \frac{6 – 3\sqrt{3}}{4 – 3}$
$= 6 – 3\sqrt{3}$
Question 2: Choose the correct option and justify your choice:
(i) $\frac{2 \tan 30^{\circ}}{1 + \tan^{2} 30^{\circ}}$
(A) $\sin 60^{\circ}$
(B) $\cos 60^{\circ}$
(C) $\tan 60^{\circ}$
(D) $\sin 30^{\circ}$
$\frac{2 \tan 30^{\circ}}{1 + \tan^{2} 30^{\circ}}$
$= \frac{2 \times \left(\frac{1}{\sqrt{3}}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right)^{2}}$
$= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$
$= \frac{\frac{2}{\sqrt{3}}}{\frac{3 + 1}{3}}$
$= \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$
$= \frac{2}{\sqrt{3}} \times \frac{3}{4}$
$= \frac{3}{2\sqrt{3}}$
$= \frac{\sqrt{3} \times \sqrt{3}}{2\sqrt{3}}$
$= \frac{\sqrt{3}}{2}$
$= \sin 60^{\circ}$
Option (A)
(ii)$\frac{1 – \tan^{2} 45^{\circ}}{1 + \tan^{2} 45^{\circ}}$
(A) $\tan 90^{\circ}$
(B) $1$
(C) $\sin 45^{\circ}$
(D) $0$
$\frac{1 – \tan^{2} 45^{\circ}}{1 + \tan^{2} 45^{\circ}}$
$= \frac{1 – (1)^{2}}{1 + (1)^{2}}$
$= \frac{1 – 1}{1 + 1}$
$= \frac{0}{2}$
$= 0$
Option (D)
(iii) $\sin 2A = 2 \sin A$ is true when $A =$
(A) $ 0^{\circ}$
(B) $ 30^{\circ}$
(C) $ 45^{\circ}$
(D) $ 60^{\circ}$
Given, $\sin 2A = 2 \sin A$
When, $A = 0^{\circ}$
$\text{LHS} = \sin(2 \times 0^{\circ}) = \sin 0^{\circ} = 0$
$\text{RHS} = 2 \sin 0^{\circ} = 2 \times 0 = 0$
$\text{LHS} = \text{RHS}$
Option (A)
(iv) $\frac{2 \tan 30^{\circ}}{1 – \tan^{2} 30^{\circ}}$
(A) $\cos 60^{\circ}$
(B) $\sin 60^{\circ}$
(C) $\tan 60^{\circ}$
(D) $\sin 30^{\circ}$
$\frac{2 \tan 30^{\circ}}{1 – \tan^{2} 30^{\circ}}$
$= \frac{2 \times \left(\frac{1}{\sqrt{3}}\right)}{1 – \left(\frac{1}{\sqrt{3}}\right)^{2}}$
$= \frac{\frac{2}{\sqrt{3}}}{1 – \frac{1}{3}}$
$= \frac{\frac{2}{\sqrt{3}}}{\frac{3 – 1}{3}}$
$= \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
$= \frac{2}{\sqrt{3}} \times \frac{3}{2}$
$= \frac{3}{\sqrt{3}}$
$= \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}$
$= \sqrt{3}$
$= \tan 60^{\circ}$
Option (C)
Question 3
(i) If $\tan(A+B) = \sqrt{3}$ and $\tan(A-B) = \frac{1}{\sqrt{3}}$; $0^{\circ} < A+B \le 90^{\circ}; A > B$, find A and B.
Solution:
Given,
$\tan(A+B) = \sqrt{3}$
$\implies \tan(A+B) = \tan 60^{\circ}$
∴ $A+B = 60^{\circ} \quad \text{— (1)}$
and
$\tan(A-B) = \frac{1}{\sqrt{3}}$
$\implies \tan(A-B) = \tan 30^{\circ}$
∴ $A-B = 30^{\circ} \quad \text{— (2)}$
Adding equation (i) and (ii), we have
$(A+B) + (A-B) = 60^{\circ} + 30^{\circ}$
$\implies A + B + A – B = 90^{\circ}$
$\implies 2A = 90^{\circ}$
$\implies A = \frac{90^{\circ}}{2}$
∴$A = 45^{\circ}$
Now, substituting the value of $A = 45^{\circ}$ in equation (i), we have
$45^{\circ} + B = 60^{\circ}$
$\implies B = 60^{\circ} – 45^{\circ}$
$\implies B = 15^{\circ}$
Therefore, $A = 45^{\circ}$ and $B = 15^{\circ}$
(ii) If $\sin(x+y) = 1$ and $\cos(x-y) = \frac{\sqrt{3}}{2}$ and $x>y$; $0^{\circ} \le x+y \le 90^{\circ}$, then find x and y.
Solution:
Given,
$\sin(x+y) = 1$
$\implies \sin(x+y) = \sin 90^{\circ}$
∴ $x+y = 90^{\circ} \quad \text{— (1)}$
and
$\cos(x-y) = \frac{\sqrt{3}}{2}$
$\implies \cos(x-y) = \cos 30^{\circ}$
∴ $x-y = 30^{\circ} \quad \text{— (2)}$
Adding equation (1) and (2), we have
$(x+y) + (x-y) = 90^{\circ} + 30^{\circ}$
$\implies x + y + x – y = 120^{\circ}$
$\implies 2x = 120^{\circ}$
$\implies x = \frac{120^{\circ}}{2}$
∴ $x = 60^{\circ}$
Now, substituting the value of $x = 60^{\circ}$ in equation (1), we have
$60^{\circ} + y = 90^{\circ}$
$\implies y = 90^{\circ} – 60^{\circ}$
$\implies y = 30^{\circ}$
Therefore, $x = 60^{\circ}$ and $y = 30^{\circ}$
Question 4
State whether the following are true or false. Justify your answer.
(i) $\sin(A+B) = \sin A + \sin B$
Solution:
Let $A = 60^{\circ}$ and $B = 30^{\circ}$
$\text{LHS} = \sin(60^{\circ}+30^{\circ})$
$\implies \text{LHS} = \sin 90^{\circ}$
∴ $\text{LHS} = 1$
$\text{RHS} = \sin 60^{\circ} + \sin 30^{\circ}$
$\implies \text{RHS} = \frac{\sqrt{3}}{2} + \frac{1}{2}$
∴ $\text{RHS} = \frac{\sqrt{3}+1}{2}$
Since $\text{LHS} \neq \text{RHS}$
Therefore, the statement is False.
(ii) The value of $\sin\theta$ increases as $\theta$ increases.
Solution:
We know from the trigonometric table:
$\sin 0^{\circ} = 0$
$\sin 30^{\circ} = \frac{1}{2} = 0.5$
$\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$
$\sin 60^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866$
$\sin 90^{\circ} = 1$
Here, the value increases from $0$ to $1$.
Therefore, the statement is True.
(iii) The value of $\cos\theta$ increases as $\theta$ increases.
Solution:
We know from the trigonometric table:
$\cos 0^{\circ} = 1$
$\cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866$
$\cos 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$
$\cos 60^{\circ} = \frac{1}{2} = 0.5$
$\cos 90^{\circ} = 0$
Here, the value decreases from $1$ to $0$.
Therefore, the statement is False.
(iv) $\sin\theta = \cos\theta$ for all values of $\theta$.
Solution:
Let $\theta = 30^{\circ}$
$\sin 30^{\circ} = \frac{1}{2}$
and
$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$
Since $\frac{1}{2} \neq \frac{\sqrt{3}}{2}$
Therefore, the statement is False.
(v) $\cot A$ is not defined for $A=0^{\circ}$.
Solution:
Given, $A = 0^{\circ}$
$\cot 0^{\circ} = \frac{\cos 0^{\circ}}{\sin 0^{\circ}}$
$\implies \cot 0^{\circ} = \frac{1}{0}$
∴ $\cot 0^{\circ} = \text{Not defined}$
Therefore, the statement is True.
Question 5
Solve for $\theta$, when $0^{\circ} < \theta < 90^{\circ}$:
(i) $2\cos^{2}\theta = 1$
Solution:
Given,
$2\cos^{2}\theta = 1$
$\implies \cos^{2}\theta = \frac{1}{2}$
$\implies \cos\theta = \sqrt{\frac{1}{2}}$
$\implies \cos\theta = \frac{1}{\sqrt{2}}$
$\implies \cos\theta = \cos 45^{\circ}$
∴ $\theta = 45^{\circ}$
(ii) $3\tan^{2}\theta – 1 = 0$
Solution:
Given,
$3\tan^{2}\theta – 1 = 0$
$\implies 3\tan^{2}\theta = 1$
$\implies \tan^{2}\theta = \frac{1}{3}$
$\implies \tan\theta = \sqrt{\frac{1}{3}}$
$\implies \tan\theta = \frac{1}{\sqrt{3}}$
$\implies \tan\theta = \tan 30^{\circ}$
∴ $\theta = 30^{\circ}$
(iii) $2\cos 3\theta = 1$
Solution:
Given,
$2\cos 3\theta = 1$
$\implies \cos 3\theta = \frac{1}{2}$
$\implies \cos 3\theta = \cos 60^{\circ}$
$\implies 3\theta = 60^{\circ}$
$\implies \theta = \frac{60^{\circ}}{3}$
∴ $\theta = 20^{\circ}$
(iv) $\sin\theta – \cos\theta = 0$
Solution:
Given,
$\sin\theta – \cos\theta = 0$
$\implies \sin\theta = \cos\theta$
$\implies \frac{\sin\theta}{\cos\theta} = 1$
$\implies \tan\theta = 1$
$\implies \tan\theta = \tan 45^{\circ}$
∴ $\theta = 45^{\circ}$
Question 6
If $\tan\theta = \frac{1}{\sqrt{3}}$ then angle $\theta$ is:
(a) $60^{\circ}$
(b) $30^{\circ}$
(c) $45^{\circ}$
(d) $90^{\circ}$
Solution:
Given,
$\tan\theta = \frac{1}{\sqrt{3}}$
$\implies \tan\theta = \tan 30^{\circ}$
∴ $\theta = 30^{\circ}$
Therefore, the correct option is (b).
Question 7
Assertion (A): The value of $\sin 45^{\circ}$ is equal to $\cos 45^{\circ}$.
Reason (R): Value of $\cos\theta$ is always equal to $\sin\theta$, when $0^{\circ} < \theta < 90^{\circ}$.
Choose the correct option from the given alternative.
(a) Both assertion (A) and reason (R) are true and reason (R) is correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not correct explanation of assertion (A)
(c) Assertion (A) is true, but reason (R) is false
(d) Assertion (A) is false, but reason (R) is true
Solution:
For Assertion (A):
$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$
and
$\cos 45^{\circ} = \frac{1}{\sqrt{2}}$
Since $\sin 45^{\circ} = \cos 45^{\circ}$, Assertion (A) is true.
For Reason (R):
Let $\theta = 30^{\circ}$
$\sin 30^{\circ} = \frac{1}{2}$
and
$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$
Since $\frac{1}{2} \neq \frac{\sqrt{3}}{2}$, Reason (R) is false. Therefore, the correct option is (c) Assertion (A) is true, but reason (R) is false.
Question 8
Which is the correct order of decreasing value of trigonometric ratios at $60^{\circ}$?
(a) $\sin 60^{\circ} > \cos 60^{\circ} > \tan 60^{\circ}$
(b) $\tan 60^{\circ} > \sin 60^{\circ} > \cos 60^{\circ}$
(c) $\tan 60^{\circ} > \cos 60^{\circ} > \sin 60^{\circ}$
(d) $\sin 60^{\circ} > \tan 60^{\circ} > \cos 60^{\circ}$
Solution:
We know the values at $60^{\circ}$:
$\tan 60^{\circ} = \sqrt{3} \approx 1.732$
$\sin 60^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866$
$\cos 60^{\circ} = \frac{1}{2} = 0.5$
Comparing the values, we have $1.732 > 0.866 > 0.5$ $\implies \tan 60^{\circ} > \sin 60^{\circ} > \cos 60^{\circ}$ Therefore, the correct option is (b).