Search any question & find its solution
Question:
Answered & Verified by Expert
If $x$ takes non-positive permissible value, then $\sin ^{-1} x=$
Options:
Solution:
1966 Upvotes
Verified Answer
The correct answer is:
$-\cos ^{-1} \sqrt{1-x^2}$
Let $\sin ^{-1} x=y$. Then $x=\sin y$
Since $-1 \leq x \leq 0$, therefore $\frac{-\pi}{2} \leq \sin ^{-1} x \leq 0$ and so $\frac{-\pi}{2} \leq y \leq 0$
We have $\cos y=\sqrt{1-\sin ^2 y}$
$\Rightarrow \cos y=\sqrt{1-x^2}$, for $0 \leq y \leq \pi$ ....(i)
Now $-\frac{\pi}{2} \leq y \leq 0 \Rightarrow \frac{\pi}{2} \geq-y \geq 0$
$\Rightarrow \cos (-y)=\sqrt{1-x^2}$
{from (i)}
$\Rightarrow-y=\cos ^{-1} \sqrt{1-x^2} \Rightarrow y=-\cos ^{-1} \sqrt{1-x^2}$
Since $-1 \leq x \leq 0$, therefore $\frac{-\pi}{2} \leq \sin ^{-1} x \leq 0$ and so $\frac{-\pi}{2} \leq y \leq 0$
We have $\cos y=\sqrt{1-\sin ^2 y}$
$\Rightarrow \cos y=\sqrt{1-x^2}$, for $0 \leq y \leq \pi$ ....(i)
Now $-\frac{\pi}{2} \leq y \leq 0 \Rightarrow \frac{\pi}{2} \geq-y \geq 0$
$\Rightarrow \cos (-y)=\sqrt{1-x^2}$
{from (i)}
$\Rightarrow-y=\cos ^{-1} \sqrt{1-x^2} \Rightarrow y=-\cos ^{-1} \sqrt{1-x^2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.