Question

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

A
tan12
B
tan1(cotA)
C
4tan1(1)
D
2tan1(2)

Solution

## The correct option is D $$4\:tan^{-1}\:\left ( 1 \right )$$Given,$$\tan^{-1}\left (\displaystyle \frac{1}{2}\tan 2A \right )+\tan^{-1}\left ( \cot A\right )+\tan^{-1}\left ( \cot ^{3}A\right )$$$$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)+\tan^{-1}\left(\dfrac{\cot A+\cot^{3}A}{1-\cot^{4}A}\right)$$$$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)+\tan^{-1}\left(\dfrac{\cot A}{1-\cot^{2}A}\right)$$$$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)+\tan^{-1}\left(\dfrac{\tan A}{\tan^{2}-1}\right)$$$$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)-\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)$$$$=0$$$$=\pi$$$$=4\left(\dfrac{\pi}{4}\right)$$$$=4\tan^{-1}\left(1\right)$$Mathematics

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