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Question: If \(x\)takes non-positive permissible value, then \(\sin ^ { - 1 } x\) =...

If xxtakes non-positive permissible value, then sin⁡−1x\sin ^ { - 1 } x =

A

cos⁡−11−x2\cos ^ { - 1 } \sqrt { 1 - x ^ { 2 } }

B

−cos⁡−11−x2- \cos ^ { - 1 } \sqrt { 1 - x ^ { 2 } }

C

cos⁡−1x2−1\cos ^ { - 1 } \sqrt { x ^ { 2 } - 1 }

D

π−cos⁡−11−x2\pi - \cos ^ { - 1 } \sqrt { 1 - x ^ { 2 } }

Answer

−cos⁡−11−x2- \cos ^ { - 1 } \sqrt { 1 - x ^ { 2 } }

Explanation

Solution

Let sin⁡−1x=y\sin ^ { - 1 } x = y Then x=sin⁡yx = \sin y

Since −π2≤sin⁡−1x≤0\frac { - \pi } { 2 } \leq \sin ^ { - 1 } x \leq 0 and so

−π2≤y≤0\frac { - \pi } { 2 } \leq y \leq 0

We have cos⁡y=1−sin⁡2y\cos y = \sqrt { 1 - \sin ^ { 2 } y }

⇒cos⁡y=1−x2\Rightarrow \cos y = \sqrt { 1 - x ^ { 2 } }, for 0≤y≤π0 \leq y \leq \pi …..(i)

Now −π2≤y≤0⇒π2≥−y≥0- \frac { \pi } { 2 } \leq y \leq 0 \Rightarrow \frac { \pi } { 2 } \geq - y \geq 0

⇒cos⁡(−y)=1−x2\Rightarrow \cos ( - y ) = \sqrt { 1 - x ^ { 2 } } {from (i)}

⇒−y=cos⁡−11−x2⇒y=−cos⁡−11−x2\Rightarrow - y = \cos ^ { - 1 } \sqrt { 1 - x ^ { 2 } } \Rightarrow y = - \cos ^ { - 1 } \sqrt { 1 - x ^ { 2 } } .