Question

If $si{n}^{-1}x+si{n}^{-1}y=\frac{2\pi }{3}$
$co{s}^{-1}x-co{s}^{-1}y=\frac{\pi }{3}$, then

• A. $x=\frac{1}{2}$
• B. $x=1$
• C. $y=\frac{1}{2}$
• D. $y=1$

Answer 1

Answer: (A), (D)

The correct options are
A $x=\frac{1}{2}$
D $y=1$

$si{n}^{-1}\left(x\right)+si{n}^{-1}\left(y\right)=\frac{2\pi }{3}$

$\frac{\pi }{2}-co{s}^{-1}\left(x\right)+\frac{\pi }{2}-co{s}^{-1}\left(y\right)=\frac{2\pi }{3}$ $[\because si{n}^{-1}(x)+co{s}^{-1}(x)=\frac{\pi }{2}]$

Hence

$co{s}^{-1}\left(x\right)+co{s}^{-1}\left(y\right)=\frac{\pi }{3}$...(i)

And

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

Adding i and ii, we get

$2co{s}^{-1}\left(x\right)=\frac{2\pi }{3}$
$co{s}^{-1}\left(x\right)=\frac{\pi }{3}$

substitute in $\left(i\right)$ we get

$co{s}^{-1}\left(y\right)=\frac{\pi }{3}-\frac{\pi }{3}=0$
$\therefore x=\cos \frac{\pi }{3}=\frac{1}{2}$ and $y=\cos 0=1$.