Question

The value of ${\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}\left(\cot A\right)+{\tan }^{-1}\left({\cot }^{3}A\right)$, for $0<A<\pi /4$, is :

• A. ${\tan }^{-1}2$
• B. ${\tan }^{-1}\left(\cot A\right)$
• C. $4ta{n}^{-1}\left(1\right)$
• D. $2{\tan }^{-1}\left(2\right)$

Answer 1

Answer: (C)

$4ta{n}^{-1}\left(1\right)$

Given,${\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}\left(\cot A\right)+{\tan }^{-1}\left({\cot }^{3}A\right)$

$={\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)+{\tan }^{-1}\left(\frac{\cot A+{\cot }^{3}A}{1-{\cot }^{4}A}\right)$

$={\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)+{\tan }^{-1}\left(\frac{\cot A}{1-{\cot }^{2}A}\right)$

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

$={\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)-{\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)$

$=0$

$=\pi$

$=4\left(\frac{\pi }{4}\right)$

$=4{\tan }^{-1}\left(1\right)$