Mark the tick against the correct answer in the following:
$\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)=?$
A. $\pi$
B. $\frac{2 \pi}{3}$
C. $\frac{3 \pi}{4}$
D. $\frac{\pi}{2}$
To Find: The value of $\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$
Now, let $x=\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$
$\Rightarrow x=\frac{\pi}{4}+\left[\pi-\cos ^{-1}\left(\frac{1}{2}\right)\right]+\left[-\sin ^{-1} \frac{1}{2}\right]\left(\because \tan \left(\frac{\pi}{4}\right)=1\right.$ and $\left.\cos ^{-1}(-\theta)=\left[\pi-\cos ^{-1} \theta\right] a n d \sin ^{-1}(-\theta)=-\sin ^{-1} \theta\right)$
$\Rightarrow x=\frac{\pi}{4}+\left[\pi-\frac{\pi}{3}\right]+\left[-\frac{\pi}{6}\right]$
$\Rightarrow x=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}$
$\Rightarrow x=\frac{3 \pi}{4}$
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