Question

Find 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 $0<A<\frac{\pi }{4}$

• A. $-\pi /2$
• B. $+\pi /2$
• C. $-\pi$
• D. $+\pi$

Answer 1

Answer: (D)

$+\pi$

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

For $0<A<\left(\pi /4\right)$, $\cot A>1$
$\Rightarrow$ $\left(\cot A\right)\left({\cot }^{3}A\right)>1$
Then, ${\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}\left(\cot A\right)+{\tan }^{-1}\left({\cot }^{3}A\right)$
$={\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)+\pi +{\tan }^{-1}\left(\frac{\cot A+{\cot }^{3}A}{1-{\cot }^{4}A}\right)$
$={\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)+\pi +{\tan }^{-1}\left(\frac{\cot A}{1-{\cot }^{2}A}\right)$
$={\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)+\pi +{\tan }^{-1}\left(\frac{\tan A}{{\tan }^{2}A-1}\right)=\pi$