\(\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right)+\tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right)\) is

\(\begin{array}{ll} & \quad \text { Let } \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)=\theta \\ \therefore \quad & \cos ^{-1}\left(\frac{a}{b}\right)=2 \theta \\ \therefore \quad & \cos 2 \theta=\frac{a}{b} \\ \therefore \quad & \tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]\end{array}\)
\(\begin{aligned}
& =\tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right) \\
& =\frac{1+\tan \theta}{1-\tan \theta}+\frac{1-\tan \theta}{1+\tan \theta} \\
& =\frac{(1+\tan \theta)^2+(1-\tan \theta)^2}{1-\tan ^2 \theta} \\
& =\frac{2\left(1+\tan ^2 \theta\right)}{1-\tan ^2 \theta} \\
& =\frac{2}{1-\tan ^2 \theta} \\
& \text {. } \overline{1+\tan ^2 \theta} \\
& =\frac{2}{\cos 2 \theta} \\
& =\frac{2}{\frac{a}{b}}=\frac{2 b}{a}
\end{aligned}\)