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  • Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 5

Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 5

Answers (1)

Answer:

                \frac{-\pi}{2}

Hint:

You must know the rules of trigonometric and inverse trigonometric function.

Given:

x< 0,\tan^{-1}x+\tan^{-1}\frac{1}{x}

Solution:

Using the property of inverse trigonometry

                \begin{gathered} \tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\tan ^{-1}\left(\frac{x+\frac{1}{x}}{1-1}\right)\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan ^{-1} a+\tan ^{-1} b=\tan ^{-1}\left(\frac{a+b}{1-a b}\right)\right] \\ \end{gathered}

                                                    =\tan ^{-1}\left(\frac{\frac{x^{2}+1}{x}}{0}\right)

Since,\left ( x+\frac{1}{x} \right )=Negative value = integer = -a

                \tan^{-1}\left ( \frac{-a}{0} \right )

  \Rightarrow \tan^{-1}x+\tan^{-1}\frac{1}{x}=\tan^{-1}\left ( -\infty \right )

Using value of inverse trigonometry,

\Rightarrow \tan^{-1}\left ( -\infty \right )=\frac{-\pi}{2}        

Substituting the value, we get

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

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