#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 1 Subquestion (iii) Maths Textbook Solution.

$\frac{\pi }{12}$
Hint:
$Separate\, and\,\, reorganize \,\, the\, values$
Given:
$Find \: principal\: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right )$
Solution:
$\sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right )$$= \sin^{-1}\left ( \frac{\sqrt{3}}{2\sqrt{2}} -\frac{1}{2\sqrt{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} -\frac{1}{2}\times \frac{1}{\sqrt{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2}\times \sqrt{1-\left ( \frac{1}{\sqrt{2}} \right )^{2}}-\frac{1}{\sqrt{2}}\times \sqrt{1-\left ( \frac{\sqrt{3}}{2} \right )^{2}} \right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right )-\sin^{-1}\left ( \frac{1}{\sqrt{2}} \right )$
$= \frac{\pi }{3}-\frac{\pi }{4}$
$=\frac{\pi }{12}$
$There\! fore, principal \: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right ) is\: \frac{\pi }{2}.$