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  • Please solve RD Sharma Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 1 Subquestion (i) maths textbook solution.

Please solve RD Sharma Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 1 Subquestion (i) maths textbook solution.

Answers (1)

Answer : \frac{\pi }{6}

Hint : The branch with range \left[\frac{-\pi}{2}, \frac{\pi}{2}\right] is called the principal value branch of function \tan^{-1}.

          \tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]

Given :  \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)

Explanation :

Let   \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=y                                                                      .....(i)

\tan y=\frac{1}{\sqrt{3}} \quad\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \left[\because \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]

\tan y=\tan \frac{\pi}{6}

\begin{aligned} &y=\frac{\pi}{6} \\ &\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6} \end{aligned}                                                        [ From equation (i) ]

The range of principal value branch of \tan^{-1} is \left[\frac{-\pi}{2}, \frac{\pi}{2}\right]

\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]

Hence, principal value of  \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}.

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