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Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 28

Answers (1)

Answer:

                \frac{1}{3}

Hints:

You must know the rules of inverse functions.

Given:

                \tan^{-1}x+\tan^{-1}\frac{1}{2}=\frac{\pi}{4}

 

Solution:

                \tan^{-1}x+\tan^{-1}\frac{1}{2}=\frac{\pi}{4}

                \tan^{-1}\left ( \frac{x+\frac{1}{2}}{1-\frac{x}{2}} \right )=\frac{\pi}{4}                                                                 \left [ \because \tan^{-1} a+\tan^{-1}b=\tan^{-1}\left ( \frac{a+b}{1-ab} \right )\right ]

                \frac{2x+1}{2}\times\frac{2}{2-x}=\tan\frac{\pi}{4}

                2x+1=\left ( 2-x \right )\left ( 1 \right )

                2x+x=2-1

                3x=1

                x=\frac{1}{3}

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