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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 2 Subquestion (ii) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 2 Subquestion (ii) Maths Textbook Solution

Answers (1)

Answer:
\frac{\pi }{6}
Hint:
\sin^{-1}x= y
\sin y= x
Given:
Find\: principal\, value \, o\! f \sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \}
Solution:
\sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \}
= \sin^{-1}\left [ \cos \left ( \frac{\pi }{3} \right ) \right ] \; \; \; \; \; \; \; \; \left [ \because \left ( \frac{\sqrt{3}}{2} \right ) = \frac{\pi }{3}\right ]
= \left ( \frac{1}{2} \right )\; \; \; \; \; \; \; \; \left [ \because \cos \frac{\pi }{3}= \frac{1}{2} \right ]

Let , \left ( \frac{1}{2} \right )= y
\sin \sin y =\frac{1}{2}
Range\, o\! f \sin^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and \, \sin \sin \left ( \frac{\pi }{6} \right ) =\frac{1}{2}
Therefore, principal \, value\, o\! f \, \sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \} is \, \frac{\pi }{6}.

 

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