Trigonometric Functions — Maths STD 12 Science — Question

$\tan ^{-1} \frac{1}{2}=\frac{1}{3} \tan ^{-1} \frac{11}{2}$

i.e. to show that $3 \tan ^{-1} \frac{1}{2}=\tan ^{-1} \frac{11}{2}$

$\mathrm{LHS}^2 3 \tan ^{-1} \frac{1}{2}$

$=2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2}$

$=\tan ^{-1}\left[\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\right]+\tan ^{-1} \frac{1}{2}$

$\cdots\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]$

$=\tan ^{-1}\left[\frac{1}{(3 / 4)}\right]+\tan ^{-1} \frac{1}{2}$

$=\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{2}$

$=\tan ^{-1}\left[\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{3} \times \frac{1}{2}}\right]$

$\begin{aligned} & =\tan ^{-1}\left(\frac{8+3}{6-4}\right) \\ & =\tan ^{-1}\left(\frac{11}{2}\right)=\text { RHS. }\end{aligned}$