If ^ -1 ( 2 a 1+a^ 2 )- ^ -1 ( 1-b^ 2 1+b^ 2 )= ^ -1 ( 2 x 1-x^ 2 ), then what is the value of x ?

Correct answer: D. (a-b)/(1+a b) Given, ^ -1 ( 2 a 1+a^ 2 )- ^ -1 ( 1-b^ 2 1+b^ 2 )= ^ -1 ( 2 x 1-x^ 2 ) 2 ^ -1 a-2 ^ -1 b=2 ^ -1 x ^ -1 a- ^ -1 b= ^ -1 x ^ -1 ((a-b)/(1+a b))= ^ -1 x x=(a-b)/(1+a b)

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

BITSAT 2016 · Mathematics · Inverse Trigonometric Functions

If ^ -1 ( 2 a 1+a^ 2 )- ^ -1 ( 1-b^ 2 1+b^ 2 )= ^ -1 ( 2 x 1-x^ 2...

\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1} \left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right), then what is the value of

?

Options

A $a / b$
B $a b$
C $b / a$
D $\frac{a-b}{1+a b}$

Correct answer

D. $\frac{a-b}{1+a b}$

Step-by-step solution

Given,

\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\therefore \quad 2 \tan ^{-1} a-2 \tan ^{-1} b=2 \tan ^{-1} x\Rightarrow \tan ^{-1} a-\tan ^{-1} b=\tan ^{-1} x\Rightarrow \quad \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x\Rightarrow \quad x=\frac{a-b}{1+a b}

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