Q48 / 125 · BITSAT 2016 · Mathematics
If $\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$,
then what is the value of $x$ ?
- A$a / b$
- B$a b$
- C$b / a$
- Correct answerD$\frac{a-b}{1+a b}$
Solution
Given,
$\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
$\therefore \quad 2 \tan ^{-1} a-2 \tan ^{-1} b=2 \tan ^{-1} x$
$\Rightarrow \tan ^{-1} a-\tan ^{-1} b=\tan ^{-1} x$
$\Rightarrow \quad \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x$
$\Rightarrow \quad x=\frac{a-b}{1+a b}$
Hence, the correct answer is option: 'D'.