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

Answer: $x= 1$
Given:$\cos^{-1}x+\sin^{-1}\frac{x}{2}-\frac{\pi }{6}=0$
Hint: $\sin^{-1}\left (\frac{1}{2} \right )= \frac{\pi }{6}$
$\sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]$
Solution: We have   $\cos^{-1}x+\sin^{-1}\frac{x}{2}-\frac{\pi }{6}=0$
$\Rightarrow \left ( \frac{\pi }{2}-\sin^{-1} x\right )+\sin^{-1}\frac{x}{2}-\frac{ \pi }{6}= 0$
$\Rightarrow \sin^{-1}\frac{x}{2}-\sin^{-1}x+\frac{\pi }{2}-\frac{ \pi }{6}= 0$
$\Rightarrow \sin^{-1}\frac{x}{2}-\sin^{-1}x+\frac{\pi }{3}= 0$
$\Rightarrow \sin^{-1}\frac{x}{2}= \sin^{-1}x-\sin^{-1}\frac{\sqrt{3}}{2}\; \; \; \; \; \; \left [ \because \sin \frac{\pi }{3}= \frac{\sqrt{3}}{2} \right ]$
Using the formula,
$\sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]$
$\Rightarrow \sin^{-1}\frac{x}{2}= \sin^{-1}\left [ x\sqrt{1-\left ( \frac{\sqrt{3}}{2} \right )^{2}}-\frac{\sqrt{3}}{2}\sqrt{1-x^{2}} \right ]$
$\Rightarrow \sin^{-1}\frac{x}{2}= \sin^{-1}\left [ x\sqrt{1-\left ( \frac{3}{4} \right )}-\frac{\sqrt{3}}{2}\sqrt{1-x^{2}} \right ]$
$\Rightarrow \frac{x}{2}= \frac{x}{2}-\frac{\sqrt{3}}{2}\sqrt{1-x^{2}}$
$\Rightarrow \sqrt{1-x^{2}}= 0$
Squaring on both sides, we get
$\Rightarrow 1-x^{2}= 0$
$\Rightarrow x= \pm 1$ [ As $x=-1$ is not satisfying the equation]
Hence $x=1$is the required answer.