If $ an ^{-1}\left(\frac{1}{4} ight)+ an ^{-1}\left(\frac{2}{9} ight)=\frac{1}{2} \cos ^{-1} x$, then $x$ is

Question

If $	an ^{-1}\left(\frac{1}{4}ight)+	an ^{-1}\left(\frac{2}{9}ight)=\frac{1}{2} \cos ^{-1} x$, then $x$ is

Aicharya · Accepted Answer

$	an ^{-1}\left(\frac{1}{4}ight)+	an ^{-1}\left(\frac{2}{9}ight)=	an ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4}ight)\left(\frac{2}{9}ight)}ight)$ $\begin{aligned} & =	an ^{-1}\left(\frac{9+8}{36-2}ight) \ & =	an ^{-1}\left(\frac{17}{34}ight) \ & =	an ^{-1}\left(\frac{1}{2}ight) \ & =\frac{1}{2} \cos ^{-1}\left[\frac{1-\left(\frac{1}{2}ight)^2}{1+\left(\frac{1}{2}ight)^2}ight]\end{aligned}$ $\cdots\left[\because 2 	an ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight)ight]$ $\begin{array}{ll} & =\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}ight) \ 	herefore \quad & x=\frac{3}{5}\end{array}$

If $\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\frac{1}{2} \cos ^{-1} x$, then…

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