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

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

Answers (1)

Answer : \frac{-\pi }{4}

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

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

Given : \tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)

Explanation :

Let y=\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)                                                                                                  ....(i)

y=\tan ^{-1}\left[2\left(\frac{-1}{2}\right)\right]                                                                                                  \left [ \because \cos\frac{2\pi }{3}=\frac{-1}{2} \right ]

\begin{aligned} &y=\tan ^{-1}(-1) \\ &\tan y=-1 \end{aligned}

\tan y=-\tan \left(\frac{\pi}{4}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \left [ \because \tan (-\theta)=-\tan \theta \right ]

y=\frac{-\pi}{4}

\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)=\frac{-\pi}{4}

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

\because \quad \tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)=\frac{-\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]

Hence, the principal value of \tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right) is \frac{-\pi}{4}.

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