Question

Let $ta{n}^{-1}y=ta{n}^{-1}x+ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)$ 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)

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