Find the value of Find the value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$

Let $\tan ^{-1}(1)=x$. Then, $\tan x=1=\tan \frac{\pi}{4}$.

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

Let $\cos ^{-1}\left(-\frac{1}{2}\right)=y$. Then, $\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)$.

$\therefore \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}$

Let $\sin ^{-1}\left(-\frac{1}{2}\right)=z .$ Then, $\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right) .$

$\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$

$\therefore \tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$

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

$=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}$