Question:
Write the value of $\sin ^{-1}\left(\frac{1}{3}\right)-\cos ^{-1}\left(-\frac{1}{3}\right)$.
Solution:
We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ and $\cos ^{-1}(-x)=\pi-\cos ^{-1} x$.
$\therefore \sin ^{-1}\left(\frac{1}{3}\right)-\cos ^{-1}\left(-\frac{1}{3}\right)=\sin ^{-1}\left(\frac{1}{3}\right)-\left[\pi-\cos ^{-1}\left(\frac{1}{3}\right)\right]$
$=\sin ^{-1}\left(\frac{1}{3}\right)-\pi+\cos ^{-1}\left(\frac{1}{3}\right)$
$=\left[\sin ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{1}{3}\right)\right]-\pi$
$=\frac{\pi}{2}-\pi \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$
$=-\frac{\pi}{2}$
$\therefore \sin ^{-1}\left(\frac{1}{3}\right)-\cos ^{-1}\left(-\frac{1}{3}\right)=-\frac{\pi}{2}$