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