Explanation
Given,$$ \tan^{-1}\left (\displaystyle \frac{1}{2}\tan 2A \right )+\tan^{-1}\left ( \cot A\right )+\tan^{-1}\left ( \cot ^{3}A\right ) $$
$$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)+\tan^{-1}\left(\dfrac{\cot A+\cot^{3}A}{1-\cot^{4}A}\right)$$
$$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)+\tan^{-1}\left(\dfrac{\cot A}{1-\cot^{2}A}\right)$$
$$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)+\tan^{-1}\left(\dfrac{\tan A}{\tan^{2}-1}\right)$$
$$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)-\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)$$
$$=0$$
$$=\pi$$
$$=4\left(\dfrac{\pi}{4}\right)$$
$$=4\tan^{-1}\left(1\right)$$
Please disable the adBlock and continue. Thank you.