Question

Let $tan^{-1}\,y=tan^{-1}\,x+tan^{-1}\left(\frac{2x}{1-x^{2}}\right) $ where $\left|x\right| < \frac{1}{\sqrt{3}}$. Then a value of $y$ is

• A. $\frac{3x-x^{3}}{1-3x^{2}}$
• B. $\frac{3x+x^{3}}{1-3x^{2}}$
• C. $\frac{3x-x^{3}}{1+3x^{2}}$
• D. $\frac{3x+x^{3}}{1+3x^{2}}$

Answer 1

Answer: (A)

$\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}$
$|x|<\frac{1}{\sqrt{3}}$
$\Rightarrow \tan ^{-1} \frac{2 x}{1-x^{2}}=2 \tan ^{-1} x$
$\Rightarrow \tan ^{-1} y=\tan ^{-1} x+2 \tan ^{-1} x$
$ =3 \tan ^{-1} x $
$ =\tan ^{-1} \frac{3 x-x^{3}}{1-3 x^{2}} $
$\Rightarrow y=\frac{3 x-x^{3}}{1-3 x^{2}}$