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

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