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

Please solve RD Sharma Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 1 Subquestion (ii) 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  y=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)                                                                                       ....(i)

       \tan y=\frac{-1}{\sqrt{3}}

      \tan y=-\tan \left(\frac{\pi}{6}\right)

      \tan y=\tan \left(\frac{-\pi}{6}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \left [ \because \tan (-\theta)=-\tan \theta \right ]

\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]

\because \quad \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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