Question

The value of ${\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}\left(cotA\right)+ta{n}^{-1}\left(co{t}^{3}A\right)$ for $0<A<\pi /4$ is

• A. $4{\tan }^{-1}\left(1\right)$
• B. $2{\tan }^{-1}\left(2\right)$
• C. 0
• D. none

Answer 1

The correct option is C 0

$ta{n}^{-1}\left(\frac{1}{2}tan2A\right)+ta{n}^{-1}\left(cotA\right)+ta{n}^{-1}\left(co{t}^{3}A\right)$
$=ta{n}^{-1}\left(\frac{tanA}{1-ta{n}^{2}A}\right)+ta{n}^{-1}\left(\frac{cotA(1+co{t}^{2}A)}{1-co{t}^{4}A}\right)$
$=ta{n}^{-1}\left(\frac{tanA}{1-ta{n}^{2}A}\right)+ta{n}^{-1}\left(\frac{cotA}{1-co{t}^{2}A}\right)$
$=ta{n}^{-1}\left(\frac{tanA}{1-ta{n}^{2}A}\right)+ta{n}^{-1}\left(\frac{tanA}{ta{n}^{2}A-1}\right)$
$=ta{n}^{-1}\left(\frac{tanA}{1-ta{n}^{2}A}\right)-ta{n}^{-1}\left(\frac{tanA}{1-ta{n}^{2}A}\right)$
$=0$