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If $3\ tan^{-1}x+cot^{-1}x=\frac{\pi}{2}$ , then $x$ equals to

Option: 1
$-\frac{\pi}{3}$

Option: 2
$0$

Option: 3
$\frac{\pi}{3}$

Option: 4
$1$

Answers (1)

best_answer

Given that,

$3\ tan^{-1}x+cot^{-1}x=\frac{\pi}{2}$

We know that,

$cot\ \theta =tan\left ( \frac{\pi}{2}-\theta \right )$

$tan^{-1}x=\frac{\pi}{2}-cot^{-1}x$

Therefore,

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

$2\ tan^{-1}x=\frac{\pi}{2}-\frac{\pi}{2}$
$tan^{-1}x=0$

$x=0$

Posted by

SANGALDEEP SINGH

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