$5\cos^{- 1}\left( \frac{1 - x^{2}}{1 + x^{2}} ight) + 7\sin^{- 1}\left( \frac{2x}{1 + x^{2}} ight) - 4 an^{- 1}\left( \frac{2x}{1 - x^{2}} ight) - an^{- 1}{x = 5\pi}$, then $x$ is equal to

Question

$5\cos^{- 1}\left( \frac{1 - x^{2}}{1 + x^{2}} ight) + 7\sin^{- 1}\left( \frac{2x}{1 + x^{2}} ight) - 4	an^{- 1}\left( \frac{2x}{1 - x^{2}} ight) - 	an^{- 1}{x = 5\pi}$, then $x$ is equal to

Dr. Abubacker · Accepted Answer

(d)
Given, $5\cos^{- 1}\left( \frac{1 - x^{2}}{1 + x^{2}} ight) + 7\sin^{- 1}\left( \frac{2x}{1 + x^{2}} ight)$
$\ \ \ \ \ \ \ \ \ \  - 4	an^{- 1}\left( \frac{2x}{1 - x^{2}} ight) - 	an^{- 1}{x = 5\pi}$
$\Rightarrow \ \ \ 5\left( 2	an^{- 1}x ight) + 7\left( 2	an^{- 1}x ight) - 4\left( 2	an^{- 1}x ight) - 	an^{- 1}x = 5\pi$
$\Rightarrow \ \ \ 15	an^{- 1}{x = 5\pi}$
$\Rightarrow \ 	an^{- 1}{x = \frac{\pi}{3}}$
$	herefore\ \ \ x = \sqrt{3}$

Q. $5\cos^{- 1}\left( \frac{1 - x^{2}}{1 + x^{2}} \right) + 7\sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right) - 4\tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right) - \tan^{- 1}{x = 5\pi}$, then $x$ is equal to

Explanation

Reviewed by Dr. Abubacker

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