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Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 25

Answers (1)

Answer:

                \frac{2}{3}

Hints:

You must know the rules of inverse trigonometric functions

Given:

                \cos\left ( 2\sin^{-1}x \right )=\frac{1}{9}

Solution:

                \cos\left ( 2\sin^{-1}x \right )=\frac{1}{9}

Let  sin^{-1}x=\theta

                 x=sin\theta

                \therefore \cos2\theta=\left ( \frac{1}{9}\right )

                1-2\sin^{2}\theta=\frac{1}{9}

                2\sin^{2}\theta=\frac{8}{9}

                x^{2}=\frac{4}{9}

               x=\pm \frac{2}{3}

 

 

 

 

Posted by

infoexpert27

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