Mark the tick against the correct answer in the following:

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 $\left.\cos ^{-1}(-\theta)=\left[\pi-\cos ^{-1} \theta\right] a n d \sin ^{-1}(-\theta)=-\sin ^{-1} \theta\right)$

$\Rightarrow x=\frac{\pi}{4}+\left[\pi-\frac{\pi}{3}\right]+\left[-\frac{\pi}{6}\right]$

$\Rightarrow x=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}$

$\Rightarrow x=\frac{3 \pi}{4}$