Question

if $tan^{-1} y = tan^{-1} x + tan^{-1} \bigg(\frac{2x}{1-x^2}\bigg), \, where \, \, |x| < \frac{1}{\sqrt 3}$
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, $tan^{-1} y = tan^{-1} x + tan^{-1} \bigg(\frac{2x}{1-x^2}\bigg)
where I x I< $\frac{1}{\sqrt 3} \, \, \Rightarrow \, \, \, tan^{-1} y=tan^{-1} $\Bigg \{ \frac{x+\frac{2x}{1-x^2}}{1-x\bigg(\frac{2x}{1-x^2}\bigg)}\Bigg \}$
$ \, \, \, \, \, \, \Bigg[\because \, \, \, tan^{-1} x+tan^{-1} y =tan^{-1} \bigg(\frac{x+y}{1-xy}\bigg)$
$ $where x > 0, y > 0 and xy < 1 $\Bigg]$
$ =tan^{-1}\bigg(\frac{x-x^2+2x}{1-x^2-2x^2}\bigg)$
$tan^{-1} y =tan^{-1}\bigg(\frac{3x-x^3}{1-3x^2}\bigg) \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 $\theta$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, -\frac{\pi}{6} < \theta < \frac{\pi}{6}$
$\therefore \, \, \, \tan^{-1} y=\theta+ tan^{-1}(tan 2\theta)=\theta+2\theta=3\theta$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, y-tan 3\theta$
$\Rightarrow \, \, \, \, \, \, \, y= \frac{3tan \theta-than^3 theta}{1-3tan^2 \theta} \Rightarrow \, \, y=\frac{3x-x^3}{1-3x^2}$