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Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 31

Answers (1)

Answer:

                \left [ -\pi,\pi \right ]

Hint:

You must know the rules of inverse trigonometric function.

Given:

                y=2\tan^{-1}x+\sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )

Solution:

                y= 2\tan^{-1}x+\sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )

 Let, x=\tan \tan\theta

 \therefore y=2\tan\theta+\left ( \frac{2\tan\tan\theta}{1+\theta} \right )                                                                                                    

Y=2\theta+\sin^{-1}\left ( \sin2\theta \right )                                                                                                \left [ \therefore \sin2\theta=\left ( \frac{2\tan\theta}{1+\tan^{2}\theta} \right ) \right ]                                                                    

\Rightarrow y=2\theta+2\theta

\Rightarrow y=4\theta                                                                                 \left [ \therefore \theta=\tan^{-1} \left ( x \right )\right ]

\Rightarrow y=4\tan^{-1}x

          \because -\frac{\pi}{2}< \tan^{-1}x< \frac{\pi}{2}

          \because -\frac{4\pi}{2}<4 \tan^{-1}x< \frac{4\pi}{2}

\Rightarrow -2\pi< 4\tan^{-1}x< 2\pi

\Rightarrow -2\pi< y< 2\pi                                                                                     \left [ y=4\tan^{-1}x \right ]

Range of y is \left [ -2\pi,2\pi \right ]

 

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