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  • Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 27

Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 27

Answers (1)

Answer:

                \frac{1}{2}

Hints:

You must know the rules and values of inverse function

Given:

\tan^{-1}x=\frac{\pi}{4}-\tan^{-1}\frac{1}{3}

Solution:

                \tan^{-1}x=\frac{\pi}{4}-\tan^{-1}\frac{1}{3}

                \tan^{-1}x+\tan^{-1}\frac{1}{3}=\frac{\pi}{4}

                \tan^{-1}\left ( \frac{x+\frac{1}{3}}{1-\frac{x}{3}} \right )=\frac{\pi}{4}                                                                  \left [ \because \tan^{-1} a+\tan^{-1}b=\tan^{-1}\left ( \frac{a+b}{1-ab} \right )\right ]                                                        

                \frac{3x+1}{3}\times\frac{3}{3-x}=\tan\frac{\pi}{4}

                \frac{3x+1}{3-x}=1

                3x+1=3-x

                3x+x=3-1

                4x=2

                x=\frac{1}{2}

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