Question

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

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

Answer 1

Answer: (D)

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