$ an^{- 1}\left( \frac{3x - x^{3}}{1 - 3x^{2}} ight) - an^{- 1}\left( \frac{2x}{1 - x^{2}} ight)$ is equal to

Question

$	an^{- 1}\left( \frac{3x - x^{3}}{1 - 3x^{2}} ight) - 	an^{- 1}\left( \frac{2x}{1 - x^{2}} ight)$ is equal to

Dr. Abubacker · Accepted Answer

(c)
Let $x = 	an{	heta\ \  \Rightarrow \ \ \ 	heta = 	an^{- 1}x}$
$	herefore\ 	an^{- 1}\left( \frac{3x - x^{3}}{1 - 3x^{2}} ight) - 	an^{- 1}\left( \frac{2x}{1 - x^{2}} ight)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 3	an^{- 1}{x - 2	an^{- 1}x}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 	an^{- 1}x$

Q. $\tan^{- 1}\left( \frac{3x - x^{3}}{1 - 3x^{2}} \right) - \tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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