If $ an^{- 1}{\left( \frac{1}{3} ight) + an^{- 1}{\left( \frac{3}{4} ight) - an^{- 1}{\left( \frac{x}{3} ight) = 0}}}$, then $x$ is equal to

Question

If $	an^{- 1}{\left( \frac{1}{3} ight) + 	an^{- 1}{\left( \frac{3}{4} ight) - 	an^{- 1}{\left( \frac{x}{3} ight) = 0}}}$, then $x$ is equal to

Dr. Abubacker · Accepted Answer

(d)
Given, $	an^{- 1}{\left( \frac{1}{3} ight) + 	an^{- 1}\left( \frac{3}{4} ight)} = 	an^{- 1}\left( \frac{x}{3} ight)$
$\Rightarrow \ 	an^{- 1}\left( \frac{\frac{1}{3} + \frac{3}{4}}{1 - \frac{1}{3} 	imes \frac{3}{4}} ight) = 	an^{- 1}\left( \frac{x}{3} ight)$
$\Rightarrow \ \frac{13}{9} = \frac{x}{3}\ \  \Rightarrow \ \ x = \frac{13}{3}$

Q. If $\tan^{- 1}{\left( \frac{1}{3} \right) + \tan^{- 1}{\left( \frac{3}{4} \right) - \tan^{- 1}{\left( \frac{x}{3} \right) = 0}}}$, then $x$ is equal to

Explanation

Reviewed by Dr. Abubacker

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