Write the value

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 $\cos ^{-1}(-x)=\pi-\cos ^{-1} x$.

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

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

$=\left[\sin ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{1}{3}\right)\right]-\pi$

$=\frac{\pi}{2}-\pi \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$

$=-\frac{\pi}{2}$

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

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now