Question

If ${\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(y\right)=\frac{2\pi }{3}$, then ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)=$

• A. $\frac{2\pi }{3}$
• B. $\frac{\mathrm{\pi }}{3}$
• C. $\frac{\mathrm{\pi }}{6}$
• D. $\mathrm{\pi }$

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{3}$

Explanation for the correct option.

Given that, ${\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(y\right)=\frac{2\pi }{3}$

Now using identity, ${\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)=\frac{\pi }{2}$ we can write:

$\begin{array}{rcl}\frac{\mathrm{\pi }}{2}-{\cos }^{-1}x+\frac{\mathrm{\pi }}{2}-{\cos }^{-1}y& =& \frac{2\mathrm{\pi }}{3}\\ -\left(c{\mathrm{os}}^{-1}x+{\cos }^{-1}y\right)+\mathrm{\pi }& =& \frac{2\mathrm{\pi }}{3}\\ \mathrm{c}{\mathrm{os}}^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{y}& =& \mathrm{\pi }-\frac{2\mathrm{\pi }}{3}\\ \mathrm{c}{\mathrm{os}}^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{y}& =& \frac{\mathrm{\pi }}{3}\end{array}$

Hence, option B is correct.