if

Question:

If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z+\sin ^{-1} t=2 \pi$, then find the value of $x^{2}+y^{2}+z^{2}+t^{2}$

Solution:

We know that the maximum value of $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ and $\sin ^{-1} t$ is $\frac{\pi}{2}$

Now,

$\mathrm{LHS}=\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z+\sin ^{-1} t$

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

$=2 \pi=\mathrm{RHS}$

Now,

$\sin ^{-1} x=\frac{\pi}{2}, \sin ^{-1} y=\frac{\pi}{2}, \sin ^{-1} z=\frac{\pi}{2}$ and $\sin ^{-1} t=\frac{\pi}{2}$

$\Rightarrow x=\sin \frac{\pi}{2}, y=\sin \frac{\pi}{2}, z=\sin \frac{\pi}{2}$ and $t=\sin \frac{\pi}{2}$

$\Rightarrow x=1, y=1, z=1$ and $t=1$

$\therefore x^{2}+y^{2}+z^{2}+t^{2}=1+1+1+1=4$

 

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