Search any question & find its solution
Question:
Answered & Verified by Expert
If $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$, then what is the value of $x$ ?
Options:
Solution:
2861 Upvotes
Verified Answer
The correct answer is:
$\mathrm{x}=\frac{\sqrt{3}}{2}$
As given:
$\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$ ...(1)
and we know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ ...(2)
On adding Eqs. (1) and (2) we get
$2 \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}=\frac{2 \pi}{3}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{3} \Rightarrow x=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$
$\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$ ...(1)
and we know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ ...(2)
On adding Eqs. (1) and (2) we get
$2 \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}=\frac{2 \pi}{3}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{3} \Rightarrow x=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$
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.