Question

if $ta{n}^{-1}y=ta{n}^{-1}x+ta{n}^{-1}\left(\frac{2x}{1-x^2}\right),where|x|<\frac{1}{\sqrt{3}}$
Then, the value of y is

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

Answer 1

Answer: (A)

Given, $ta{n}^{-1}y=ta{n}^{-1}x+ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)<br/>whereIxI<$\frac{1}{\sqrt 3} {\,} {\,} \Rightarrow {\,} {\,} {\,} tan^{-1} y=tan^{-1} $\left\{\frac{x+\frac{2x}{1-x^2}}{1-x\left(\frac{2x}{1-x^2}\right)}\right\}$
$[\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}(\frac{x+y}{1-xy})$
where x > 0, y > 0 and xy < 1 $]$
$=ta{n}^{-1}\left(\frac{x-x^2+2x}{1-x^2-2x^2}\right)$
$ta{n}^{-1}y=ta{n}^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)\Rightarrow y=\frac{3x-x^2}{1-3x^2}$
Alternate Solution
$|x|<\frac{1}{\sqrt{3}}\Rightarrow -\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$
Let a: = tan $\theta$
$\Rightarrow -\frac{\pi }{6}<\theta <\frac{\pi }{6}$
$\therefore {\tan }^{-1}y=\theta +ta{n}^{-1}(tan2\theta )=\theta +2\theta =3\theta$
$\Rightarrow y-tan3\theta$
$\Rightarrow y=\frac{3tan\theta -tha{n}^{3}theta}{1-3ta{n}^2\theta }\Rightarrow y=\frac{3x-x^3}{1-3x^2}$