Latest Articles

NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry Exercise 8.2

NCERT Solutions for Class 10 Maths Introduction to Trigonometry Exercise 8.2

In this page we have NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry for EXERCISE 8.2 . Hope you like them and do not forget to like , social_share and comment at the end of the page.

Formula to Remember
NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry Exercise 8.2

Question 1
Evaluate the following :
(i) $ \sin 60^{\circ} \cos 30^{\circ} + \sin 30^{\circ} \cos 60^{\circ}$
(ii) $2 \tan^2 45^{\circ} + \cos^2 30^{\circ} - sin^2 60^{\circ}$
(iii)$ \frac {\cos 45^{\circ}}{\sec 30^{\circ} + \csc 30^{\circ}}$
(iv) $ \frac {\sin 30^{\circ} + \tan 45^{\circ} - \csc 60^{\circ}}{\sec 30^{\circ} + \cos 60^{\circ} + cot 45^{\circ}}$
(v) $ \frac {5 \cos^2 60^{\circ}+ 4 \sec^2 30^{\circ}- \tan^2 45^{\circ}}{\sin^2 30^{\circ} + \cos^2 30^{\circ}}$
Solution
(i) $ \sin 60^{\circ} \cos 30^{\circ} + \sin 30^{\circ} \cos 60^{\circ}$
$= ( \frac {\sqrt {3}}{2} \times \frac {\sqrt {3}}{2})+ ( \frac {1}{2} \times \frac {1}{2}) = \frac {3}{4}+ \frac {1}{4} = \frac {4}{4} = 1$
(ii) $2 \tan^2 45^{\circ} + \cos^2 30^{\circ} - sin^2 60^{\circ}$
$= 2 \times (1)^2 + (\frac {\sqrt {3}}{2})^2 - (\frac {\sqrt {3}}{2})^2= 2$

(iii) $ \frac {\cos 45^{\circ}}{\sec 30^{\circ} + \csc 30^{\circ}}$

$= \frac {\frac {1}{\sqrt {2}}}{ \frac {2}{\sqrt {3}}+ 2} $
$= \frac {\sqrt {3}}{\sqrt {2} (2+2\sqrt {3})} = \frac {\sqrt {3}}{2 \sqrt {2} +2\sqrt {6}}$
$= \frac {\sqrt {3} (2\sqrt {6}-2 \sqrt {2})}{(2 \sqrt {2} +2\sqrt {6})(2\sqrt {6} -2 \sqrt {2})}$
$= \frac {\sqrt {3} (2\sqrt {6}-2 \sqrt {2})}{24-8} = \frac {2 \sqrt {3}(\sqrt {6}- \sqrt {2})}{16}$
$= \frac {\sqrt {3} (\sqrt {6}- \sqrt {2})}{8} = \frac {3 \sqrt {2}- \sqrt {6}}{8}$

(iv) $ \frac {\sin 30^{\circ} + \tan 45^{\circ} - \csc 60^{\circ}}{\sec 30^{\circ} + \cos 60^{\circ} + cot 45^{\circ}}$
$= \frac {1/2 + 1- 2/\sqrt {3}}{2 /\sqrt {3} + 1/2 + 1}$
$= \frac {3/2 - 2/\sqrt {3}}{3/2 + 2/\sqrt {3}}$
$ = \frac {3 \sqrt {3}-4}{3 \sqrt {3} +4}$
Similifying the fraction
$= \frac {(3 \sqrt {3}-4)(3 \sqrt {3}-4)}{(3 \sqrt {3} +4)(3 \sqrt {3}-4)}$
$= \frac {27+16 - 24 \sqrt {3}}{27-16}$
$= \frac {43-24 \sqrt {3}}{11}$

(v) $ \frac {5 \cos^2 60^{\circ}+ 4 \sec^2 30^{\circ}- \tan^2 45^{\circ}}{\sin^2 30^{\circ} + \cos^2 30^{\circ}}$
$= \frac {5(1/2)^2+4(2/\sqrt {3})^2-1^2}{(1/2)^2+(\sqrt {3}/2)^2}$
$= \frac {5/4 + 16/3 -1}{1/4 + 3/4}$
$= \frac {15+64-12}{12}$
$= \frac {67}{12}$

Question 2
Choose the correct option and justify your choice :
(i) $\frac {2 \tan 30^{\circ}}{1+ \tan^2 30^{\circ}} =$
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) $ \frac {1- \tan^2 45^{\circ}}{1+ \tan^2 45^{\circ}} =$
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0

(iii) $\sin 2A = 2 \sin A$ is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°
(iv) $\frac {2 \tan30^{\circ}}{1- \tan^2 30^{\circ}} =$
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution
(i)
$\frac {2 \tan 30^{\circ}}{1+ \tan^2 30^{\circ}} = \frac {2( \frac {1}{\sqrt {3}})}{1+(\frac {1}{\sqrt {3}})^2}$
$= \frac {6}{4 \sqrt {3}}= \frac {\sqrt {3}}{2} = \sin 60^{\circ}$
Hence (A) is correct.

(ii)
$ \frac {1- \tan^2 45^{\circ}}{1+ \tan^2 45^{\circ}} = \frac {1-1^2}{1+1^2}=0$
Hence (D) is correct.

(iii)
$\sin 2A = 2 \sin A$
We know that only value satisfies this equation from option is 0 Hence (A) is correct.

(iv)
$\frac {2 \tan30^{\circ}}{1- \tan^2 30^{\circ}} = \frac {2(\frac {1}{\sqrt {3}})}{1-(\frac {1}{\sqrt {3}})^2}$
$= \sqrt {3}= \tan 60^{\circ}$
Hence (C) is correct.

Question 3
If $\tan (A + B) = \sqrt {3} $ and $ \tan (A - B) = \frac {1}{\sqrt {3}}$ , $0^{\circ} < A + B \leq 90^{\circ}$ , A > B, find A and B.
Solution
$\tan (A + B) = \sqrt {3} $
Now we know that $\tan 60^{\circ} =\sqrt {3} $ Therefore
$ \tan (A + B) = \tan 60^{\circ}$
or $ (A + B) = 60^{\circ}$ --- (i)
Also
$ \tan (A - B) = \frac {1}{\sqrt {3}}$
Now we know that $\tan 30^{\circ} =\frac {1}{\sqrt {3}} $ Therefore
$ \tan (A - B) = \tan 30^{\circ}$
or $(A - B) = 30^{\circ}$ -- (ii)
Adding (i) and (ii), we get
$2A = 60^{\circ} + 30^{\circ}$
$A= 45^{\circ}$
Putting the value of A in equation (i)
$45^{\circ} + B = 60^{\circ}$
$B = 15^{\circ}$

Question 4
State whether the following are true or false. Justify your answer.
(i) $\sin (A + B) = \sin A + \sin B$.
(ii) The value of $\sin \theta$ increases as $\theta$ increases.
(iii) The value of $\cos \theta$ increases as $\theta$ increases.
(iv) $ \sin \theta = \cos \theta$ for all values of $\theta$.
(v) $\cot A$ is not defined for A = 0°.
Solution
(i) False.
Let A = 30° and B = 60°, then
$\sin (A + B) = \sin (30^{\circ} + 60^{\circ})= \sin 90^{\circ}= 1$ and,
$\sin A + \sin B = \sin 30^{\circ} + \sin 60^{\circ}$
$= \frac {1}{2} + \frac {\sqrt {3}}{2} = \frac {1+ \sqrt {3}}{2}$

(ii) True.
sin 0° = 0 < sin 30° = 1/2 < sin 45° = 1/√2 < sin60° =√3/2 < sin 90° = 1
Thus the value of sin θ increases as θ increases.

(iii) False.
cos 0° = 1 > cos 30° = √3/2 > cos 45° = 1/√2 > cos 60° =1/2 > cos 90° = 0
Thus the value of cos θ decreases as θ increases.

(iv) True.
We know that
$\cot A = \frac {\cos A}{\sin A}$
Therefore
$\cot 0^{\circ}= \frac {\cos 0^{\circ}}{\sin 0^{\circ}} =\frac { 1}{0} = \text {undefined}$.

Also Read