Solveeit Logo
Login

Question

Question: If \(\cos ^ { - 1 } x + \cos ^ { - 1 } y = 2 \pi\)then \(\sin ^ { - 1 } x + \sin ^ { - 1 } y\) is e...

If cos⁡−1x+cos⁡−1y=2π\cos ^ { - 1 } x + \cos ^ { - 1 } y = 2 \pithen sin⁡−1x+sin⁡−1y\sin ^ { - 1 } x + \sin ^ { - 1 } y is equal to.

A

π\pi

B

−π- \pi

C

π2\frac { \pi } { 2 }

D

None of these

Answer

−π- \pi

Explanation

Solution

cos⁡−1x+cos⁡−1y=2π\cos ^ { - 1 } x + \cos ^ { - 1 } y = 2 \pi

⇒π2−sin⁡−1x+π2−sin⁡−1y=2π\Rightarrow \frac { \pi } { 2 } - \sin ^ { - 1 } x + \frac { \pi } { 2 } - \sin ^ { - 1 } y = 2 \pi

⇒π−(sin⁡−1x+sin⁡−1y)=2π\Rightarrow \pi - \left( \sin ^ { - 1 } x + \sin ^ { - 1 } y \right) = 2 \pi

⇒sin⁡−1x+sin⁡−1y=−π\Rightarrow \sin ^ { - 1 } x + \sin ^ { - 1 } y = - \pi