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Model Paper 7 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 72 Marks
Question
Find the value of $\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left[\sin \left(\frac{-\pi}{2}\right)\right]$.
✓

Answer

We have, $\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left[\sin \left(\frac{-\pi}{2}\right)\right]$.
$=\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cot ^{-1}\left(\cot \frac{\pi}{3}\right)+\tan ^{-1}(-1)$
$=\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{6}\right)\right]+\cot ^{-1}\left[\cot \left(\frac{\pi}{3}\right)\right]+\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{4}\right)\right]$
$=\tan ^{-1}\left(-\tan \frac{\pi}{6}\right)+\cot ^{-1}\left(\cot \frac{\pi}{3}\right)+\tan ^{-1}\left(-\tan \frac{\pi}{4}\right)\left[\begin{array}{c}\because \tan ^{-1}(\tan x)=x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ \cot ^{-1}(\cot x)=x, x \in(0, \pi) \\ \text { and } \tan ^{-1}(-x)=-\tan ^{-1} x\end{array}\right]$
$=-\frac{\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4}=\frac{-2 \pi+4 \pi-3 \pi}{12}$
$=\frac{-5 \pi+4 \pi}{12}=\frac{-\pi}{12}$

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