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Chapter 2 Inverse Trigonometric Functions Exercise 2.2

In this page we have Class 12 Maths NCERT Solutions Chapter 2 Inverse Trigonometric Functions for EXERCISE 2.2 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Important Formula’s
Class 12 Maths NCERT Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2.2

Prove the following

Question 1. $3 \sin ^{-1} = \sin ^{-1}(3x - 4x^{3}) , \; x \in \left [ -\frac{1}{2}, \frac{1}{2} \right ]$
Solution
Let sin^-1x = y, then x = sin y
We get,
RHS = $\sin ^{-1} (3x - 4x^{3 }) = \sin ^{-1} (3 \sin y - 4 \sin^{3} y )\\$
$= \\\sin ^{-1} (\sin 3 y) = 3 y = 3 \sin^{-1}x$
= LHS

Question 2. $3 \cos ^{-1} x = cos ^{-1}(4x^{3} - 3x), x \in \left [ \frac{1}{2}, 1 \right ]$
Solution
Let cos^-1 x = y, then x = cos y
We get,
RHS = $\cos ^{-1} (4x^{3} - 3x) = cos^{-1}(4cos^{3} y - 3cos y)$
$= \\\cos ^{-1} (cos 3 y ) = 3 y = 3 cos^{-1} x$
= LHS

Question 3. Show that $tan ^{-1} \frac{2}{11} + \tan ^{-1} \frac{7}{24} = \tan ^{-1} \frac{1}{2}$
Solution
LHS = $tan ^{-1} \frac{2}{11} + \tan ^{-1} \frac{7}{24} $
$= tan ^{-1} \left ( \frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}} \right ) = \tan^{-1} \left (\frac{\frac{48 + 77}{11 \times 24}}{\frac{11 \times 24 - 14}{11 \times 24}} \right )\\$
$\\= tan ^{-1} \frac{48 + 77}{264 - 14} = \tan^{-1} \frac{125}{251} = \tan^{-1} \frac{1}{2}$ = RHS

Question 4. Show that $2 \tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}$
Solution
LHS = $2 \tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} $
$= \tan^{-1} \left [ \frac{2 \times \frac{1}{2}}{1 - \left ( \frac{1}{2} \right )^{2}} \right ] + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{1}{\left ( \frac{3}{4} \right )} + \tan^{-1} \frac{1}{7}\\$
$\\= \tan^{-1} \frac{4}{3} + \tan^{-1} \frac{1}{7} = \tan^{-1}\left ( \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \times \frac{1}{7}} \right )\\$
$\\= \tan^{-1} \left ( \frac{\frac{28 + 3}{3 \times 7}}{\frac{3 \times 7 -4}{3 \times 7}} \right ) = \tan^{-1} \frac{28 + 3}{21 - 4} = tan^{-1} \frac{31}{17} = RHS$

Write the following functions in the simplest form

Question 5. $\tan^{-1}\frac{\sqrt{1 + x^{2}} - 1}{x}, \; x \neq 0$
Solution
Let x = tan y
$ = \tan^{-1}\frac{\sqrt{1 + x^{2}} - 1}{a}$ = $\tan^{-1} \frac{\sqrt{1 + \tan^{2}y } - 1}{\tan y } \\$
$\\ = \tan^{-1} \left ( \frac{ \sec y - 1 }{\tan y } \right ) = \tan^{-1} \left ( \frac{1 - \cos y }{\sin y } \right )\\$
$\\\tan^{-1} \left ( \frac{2\sin^{2}\frac{y }{2}}{2\sin\frac{y }{2}\cos\frac{y }{2}}\right ) = \tan^{-1}\left ( \tan \frac{y }{2} \right )\\$
$\\= \frac{y }{2} = \frac{1}{2}\tan^{-1}x$

Question 6. $\tan^{-1}\frac{1}{\sqrt{x^{2}-1}}$, |x|> 1
Solution:
Let x = csc y
$\tan^{-1}\frac{1}{\sqrt{x^{2}-1}} = \tan^{-1}\frac{1}{\sqrt{\csc^{2}y -1}} $
$=\tan^{-1}\frac{1}{ \cot y } = \tan^{-1} \tan y = y = \csc ^{-1}x $
$= \frac{\pi}{2} - sec^{-1}x$

Question 7. $\tan^{-1} \left ( \sqrt{\frac{1 - \cos x}{1 + \cos x}} \right ), x < \pi,$
Solution
$\tan^{-1} \left ( \sqrt{\frac{1 - \cos x}{1 + \cos x}} \right ) = \tan^{-1} \left ( \sqrt{\frac{2 \sin^{2}\frac{x}{2}}{2 \cos^{2}\frac{x}{2}}} \right ) \\$
$=\\\tan^{-1} \left ( \sqrt{\tan^{2}\frac{x}{2}} \right ) = \tan^{-1}\left ( \tan \frac{x}{2} \right ) = \frac{x}{2}$

Question 8. $\tan^{-1} \left ( \frac{\cos x - \sin x}{\cos x + \sin x} \right ), 0 < x < \pi$
Solution
$ \tan^{-1} \left ( \frac{\cos x - \sin x}{\cos x + \sin x} \right ) = \tan^{-1} \left ( \frac{1 - \frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}} \right ) = \tan^{-1} \left ( \frac{1 - \tan x}{1 + \tan x} \right )\\$
$=\\\tan^{-1} \left ( \frac{1 - \tan x}{1 + 1.\tan x} \right ) = \tan^{-1} \left ( \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4}.\tan x}\right )\\$
$=\\\tan^{-1} \left [ \tan \left ( \frac{\pi}{4} - x \right )\right ] = \frac{\pi}{4} - x$

Question 9: $\tan^{-1} \frac{a}{\sqrt{x^{2} - a^{2}}}, \left | a \right | < x$
Solution
Let a = x sin y
$\tan^{-1} \frac{a}{\sqrt{x^{2} - a^{2}}} = \tan^{-1} \left ( \frac{x\sin y }{\sqrt{x^{2} - x^{2}\sin^{2}y }} \right ) = \tan^{-1}\left ( \frac{x\sin y }{x \sqrt{1 - \sin^{2}y }} \right ) \\$
$=\\\tan^{-1} \left ( \frac{x \sin y }{x \sin y} \right ) = tan ^{-1} (\tan y ) = y = \sin ^{-1} \frac{a}{x}$

Question 10.: $\tan^{-1} \left ( \frac{3x^{2}a - a^{3}}{x^{3} - 3xa^{2}} \right ) , x > 0; \frac{-x}{\sqrt{3}} \leq a\frac{x}{\sqrt{3}}$
Solution
Let a = x tan y
$\tan^{-1} \left ( \frac{3x^{2}a - a^{3}}{x^{3} - 3xa^{2}} \right ) = \tan^{-1} \left ( \frac{3x^{2}.x \tan y - x^{3}\tan^{3}y }{x^{3} - 3x.x^{2}\tan^{2}y } \right ) \\$
$=\\\tan^{-1} \left ( \frac{3x^{3} \tan y - x^{3}\tan^{3}y }{x^{3} - 3x^{3}\tan^{2}y } \right ) = \tan^{-1} \left ( \frac{3 \tan y - \tan^{3}y}{1 - 3\tan^{2}y } \right ) \\$
$=\tan^{-1} \left ( \tan 3 y \right ) = 3 y = 3 tan ^{-1} \frac{a}{x}$

Find the values of each of the following

Question 11. Solve $\tan^{-1}\left [ 2\cos \left ( 2 \sin^{-1} \frac{1}{2} \right ) \right ]$
Solution
$\tan^{-1}\left [ 2\cos \left ( 2 \sin^{-1} \frac{1}{2} \right ) \right ] = \tan^{-1}\left [ 2\cos \left ( 2 \sin^{-1} \left ( \sin \frac{\pi}{6} \right ) \right ) \right ] \\$
$=\\\tan^{-1}\left [ 2\cos \left ( 2 \times \frac{\pi}{6} \right ) \right ] = \tan^{-1} \left [ 2 \cos \left ( \frac{\pi}{3} \right ) \right ] = \tan^{-1} \left [ 2 \times \frac{1}{2} \right ]\\$
$=\tan^{-1}\left [ 1 \right ] = \frac{\pi}{4}$

Question 12. Solve $\cot \left (\tan^{-1} a + \cot ^{-1} a \right )$
Solution
$\cot \left (\tan^{-1} a + \cot ^{-1} a \right ) = \cot \left( \frac{\pi}{2} \right)$ = 0

Find the values of each of the expressions in Exercises 16 to 18.

Question 16 sin^-1(sin 2π/3)
Solution
sin^-1 (sin 2π/3)
Now We know that sin^-1 (sin x) = x if x in [π/2, -π/2], which is the principal value branch of sin^-1 x.
Now 2π/3 is not in [π/2, -π/2],So
= sin^-1 [(sin π - 2π/3)]
= sin^-1 [sin (π/3)]
= π/3

Question 17 tan^-1 (tan 3π/4)
Solution
tan^-1(tan 3π/4)
= tan^-1 [tan (3π/4)]
= tan^-1 [tan (π - π/4)]
= tan^-1 [-tan (π/4)]
= tan^-1 [ -tan (π/4)]
= -π/4

Question 18 tan [sin^-1 (3/5) + cot^-1 (3/2) ]
Solution
let x=sin^-1 (3/5), sinx= 3/5 , cos x=(1-sin²x)^1/2= 4/5
So tanx=3/4
or x=tan^-1 (3/4)
Now cot^-1 (3/2) = tan^-1 (2/3)
So
tan [sin^-1 (3/5) + cot^-1 (3/2) ]
=tan [tan^-1 (2/3) + tan^-1 (2/3) ]
=tan [tan^-1 (17/6)] =17/6

Question 19 cos^-1 [cos (7π/6)] is equal to
(A) 7π/6
(B) 5π/6
(C) π/3
(D) π/6
Solution
= cos^-1 [cos (2π - 5π/6)]
= cos^-1 [cos (5π/6)]
= 5π/6
Hence the correct option is (B)

Question 20 sin [(π/3) - sin^-1 (-1/2)] is equal to
(A) 1/2
(B) 1/3
(C) 1/4
(D) 1
Solution
sin [(π/3) - sin^-1 (-1/2)]
=sin [(π/3) - sin^-1 sin (-π/6)]
=sin [(π/3) +(π/6) ]
=sin (π/2)
=1
Hence the correct option is (D)

Question 21 tan^-1 √3 - cot^-1 (-√3) is equal to
(A) π
(B) -π/2
(C) 0
(D) 2√3
Solution
tan^-1 √3 =x
or tan (x) =√3
or x= π/3( Principle Range)

cot^-1 (-√3) =y
or cot y= -√3
or y = 5π/6 ( Principle Range)

tan^-1 √3 - cot^-1 (-√3)
= π/3 - 5π/6
=-π/2
Hence the correct option is (B)