If tan ^{-1}left(frac{1}{4}right)+tan ^{-1}le | vedclass

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(C) We use the formula $	an ^{-1} a + 	an ^{-1} b = 	an ^{-1} \left( \frac{a+b}{1-ab} ight)$.$	an ^{-1}\left(\frac{1}{4}ight)+	an ^{-1}\left(

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$\frac{3}{5}$

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(C) We use the formula $	an ^{-1} a + 	an ^{-1} b = 	an ^{-1} \left( \frac{a+b}{1-ab} ight)$.$	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)$$= 	an ^{-1}\left(\frac{\frac{9+8}{36}}{\frac{36-2}{36}}ight) = 	an ^{-1}\left(\frac{17}{34}ight) = 	an ^{-1}\left(\frac{1}{2}ight)$Now,we use the identity $2 	an ^{-1} 	heta = \cos ^{-1} \left( \frac{1-	heta^2}{1+	heta^2} ight)$,which implies $	an ^{-1} 	heta = \frac{1}{2} \cos ^{-1} \left( \frac{1-	heta^2}{1+	heta^2} ight)$.Setting $	heta = \frac{1}{2}$,we get $	an ^{-1}\left(\frac{1}{2}ight) = \frac{1}{2} \cos ^{-1} \left( \frac{1-(1/2)^2}{1+(1/2)^2} ight)$$= \frac{1}{2} \cos ^{-1} \left( \frac{1-1/4}{1+1/4} ight) = \frac{1}{2} \cos ^{-1} \left( \frac{3/4}{5/4} ight) = \frac{1}{2} \cos ^{-1} \left( \frac{3}{5} ight)$Comparing this with $\frac{1}{2} \cos ^{-1} x$,we find $x = \frac{3}{5}$.

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

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