^ -1 (1) + ^ -1 (1 2 ) + ^ -1 (1 3 ) = - Vidyadip

MCQ

$\cot^{-1}(1) + \cot^{-1} (\frac{1}{2}) + \cot^{-1}(\frac{1}{3}) =$

A
$0$
B
$\frac{3 \pi}{4}$
C
$\frac{2 \pi}{3}$
✓
$\pi$

✓

Answer

Correct option: D.

$\pi$

d
$\cot^{-1}(1) + \cot^{-1} (\frac{1}{2}) + \cot^{-1}(\frac{1}{3})$

$=\tan ^{-1}(1)+\tan ^{-1}(2)+\tan ^{-1}(3)$

$=\frac{\pi}{4}+\left[\frac{\pi}{2}-\cot ^{-1} 2\right]+\frac{\pi}{2}-\cot ^{-1} 3$

$=\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}-\left[\cot ^{-1}(2)+\cot ^{-1}(3)\right]$

$=\frac{5 \pi}{4}-\left[\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)\right]$

$=\frac{5 \pi}{4}-\left[\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right)\right]$

$=\frac{5 \pi}{4}-\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)$

$=\frac{5 \pi}{4}-\tan ^{-1}(1)$

$=\frac{5 \pi}{4}-\frac{\pi}{4}$

$=\pi$

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