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

Answers (1)

Answer:
\frac{\pi }{2}
Hint:
\tan \left ( \pi +x \right )= \tan x

Given:
Find\: principal\: value\: o\! f \sin^{-1}\left ( \tan \left ( \frac{5\pi }{4} \right ) \right )

Solution:
Let , y= \sin^{-1}\left ( \tan \left ( \frac{5\pi }{4} \right ) \right )
\sin y= \tan \left ( \frac{5\pi }{4} \right )
\sin y= \tan \left (\pi + \frac{\pi }{4} \right )
= \tan \left ( \frac{\pi }{4} \right )
=1
= \sin \left ( \frac{\pi }{2} \right )
Range\: Principal\: value\: o\! f \sin^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and \: \sin \left ( \frac{\pi }{2} \right )= \tan \left ( \frac{5\pi }{4} \right )
There\! f\! ore, principal\: value\: o\! f \sin^{-1}\left ( \tan \left ( \frac{5\pi }{4} \right ) \right ) is \: \frac{\pi }{2}.

 

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