Question

Let $ta{n}^{-1}y=ta{n}^{-1}x+ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)$ where $|x|<\frac{1}{\sqrt{3}}$. 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

The correct option is A $\frac{3x-x^3}{1-3x^2}$

Let ${\tan }^{-1}x=\theta$

$\tan 2\theta =\left(\frac{2x}{1-x^2}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=2\theta$

Then, ${\tan }^{-1}y={\tan }^{-1}x+2{\tan }^{-1}x$

${\tan }^{-1}y=3{\tan }^{-1}x$

$\Rightarrow {\tan }^{-1}y={\tan }^{-1}\frac{3x-x^3}{1-3x^2}$

$\Rightarrow y=\frac{3x-x^3}{1-3x^2}$

Hence, option A is correct.