Search any question & find its solution
Question:
Answered & Verified by Expert
$\text { If } \sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3}=\sin ^{-1} x \text {, }$ then $x$ is equal to
Options:
Solution:
2096 Upvotes
Verified Answer
The correct answer is:
$\frac{\sqrt{5}+4 \sqrt{2}}{9}$
$\begin{aligned} & \sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3} \\ & =\sin ^{-1}\left[\frac{1}{3} \sqrt{1-\frac{4}{9}}+\frac{2}{3} \sqrt{1-\frac{1}{9}}\right]=\sin ^{-1}\left[\frac{\sqrt{5}+4 \sqrt{2}}{9}\right] \text {, as }\left\{\sin ^{-1}(x)+\sin ^{-1}(y)=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\right\} \\ & \text {Therefore } \quad x=\frac{\sqrt{5}+4 \sqrt{2}}{9}\end{aligned}$
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.