#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 3 Subquestion (iii) Maths Textbook Solution.

Given: $\sin^{-1}\frac{5}{x}+\sin^{-1}\frac{12}{x}= \frac{\pi }{2}$
Hint: Here we will use $\frac{\pi }{2}-\sin^{-1}x= \cos^{-1}x$
Solution: Here we have
$\sin^{-1}\frac{5}{x}+\sin^{-1}\frac{12}{x}= \frac{\pi }{2}$
$\sin^{-1}\frac{5}{x}= \frac{\pi }{2}-\sin^{-1}\frac{12}{x}$
$\Rightarrow \sin^{-1}\frac{5}{x}= \cos^{-1}\frac{12}{x} \; \; \; \; \left [ \because \frac{\pi }{2}-\sin^{-1}x= \cos^{-1}x \right ]$
$\Rightarrow \sin^{-1}\frac{5}{x}=\sin^{-1}\left ( \sqrt{1-\left ( \frac{12}{x} \right )^{2}} \right ) \; \; \; \; \; \; \; \; \; \; [\cos^{-1}x=\sin^{-1}\sqrt{1-x^{2}}]$
$\Rightarrow \left ( \frac{5}{x}\right )^{2} =\sqrt{1-\left ( \frac{12}{x} \right )^{2}}$
Squaring on both sides, we get
$\Rightarrow \left ( \frac{5}{x}\right )^{2} =1-\left ( \frac{12}{x} \right )^{2}$
$\Rightarrow \left ( \frac{25}{x^{2}} \right )+\left ( \frac{144}{x^{2}} \right )= 1$
$\Rightarrow x^{2}= 169$
$\Rightarrow x=\pm 13$ [ As -13 is not satisfying the equation]
Hence $x= 13$ is the result