Question

If ${\sin }^{-1}x+{\sin }^{-1}y=\frac{2\pi }{3}$
and ${\cos }^{-1}x-{\cos }^{-1}y=\frac{\pi }{3}$
Then, $(x,y)$ is equal to

• A. $(0,1)$
• B. $(\frac{1}{2},1)$
• C. $(1,\frac{1}{2})$
• D. $(\frac{\sqrt{3}}{2},1)$

Answer 1

The correct option is B $(\frac{1}{2},1)$

We have,
${\sin }^{-1}x+{\sin }^{-1}y=\frac{2\pi }{3}....\left(i\right)$

${\cos }^{-1}x-{\cos }^{-1}y=\frac{\pi }{3}.....\left(ii\right)$

$(\frac{\pi }{2}-{\sin }^{-1}x)-(\frac{\pi }{2}-{\sin }^{-1}y)=\frac{\pi }{3}$ ...... $[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}]$
$\Rightarrow -{\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{3}.....\left(iii\right)$
On adding Eqns. $\left(i\right)$ and $\left(iii\right),$ we get
${\sin }^{-1}y=\frac{\pi }{2}\Rightarrow y=1$
On subtracting Eq. $\left(i\right)$ from Eq. $\left(iii\right),$ we get
${\sin }^{-1}x=\frac{\pi }{6}\Rightarrow x=\frac{1}{2}$
$\therefore (x,y)=(\frac{1}{2},1)$.