The value of

Question:

The value of $\cos ^{-1}\left(\cos \frac{5 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)$ is

(a) $\frac{\pi}{2}$

(b) $\frac{5 \pi}{3}$

(C) $\frac{10 \pi}{3}$

(d) 0

Solution:

(d) 0

We have

$\cos ^{-1}\left(\cos \frac{5 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)=\cos ^{-1}\left\{\cos \left(2 \pi-\frac{\pi}{3}\right)\right\}+\sin ^{-1}\left\{\sin \left(2 \pi-\frac{\pi}{3}\right)\right\}$

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

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

$=\frac{\pi}{3}-\frac{\pi}{3}$

$=0$

cos1(cos5π3)+sin1(sin5π3)=cos1{cos(2ππ3)}+sin1{sin(2ππ3)}=cos1{cos(π3)}+sin1{sin(π3)}=cos1{cos(π3)}sin1{sin(π3)}=π3π3=0

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