Question:
Mark the tick against the correct answer in the following:
$\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=?$
A. $\frac{\pi}{2}$
B. $\pi$
C. $\frac{3 \pi}{2}$
D. $\frac{2 \pi}{3}$
Solution:
To Find: The value of $\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$
Let, $x=\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$
$\Rightarrow \mathrm{x}=-\tan ^{-1}(1)+\left(\pi-\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)$
$\left(\because \tan ^{-1}(-\theta)=-\tan ^{-1}(\theta)\right.$ and $\left.\cos ^{-1}(-\theta)=\pi-\cos ^{-1}(\theta)\right)$
$\Rightarrow \mathrm{x}=-\frac{\pi}{4}+\left(\pi-\frac{\pi}{4}\right)$
$\Rightarrow \mathrm{x}=-\frac{\pi}{4}+\frac{3 \pi}{4}$
$\Rightarrow \mathrm{x}=\frac{\pi}{2}$