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  • Provide solution for RD Sharma maths Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 2 Sub question (ii) maths textbook solution

Provide solution for RD Sharma maths Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 2 Sub question (ii) maths textbook solution.

Answers (1)

Answer : \frac{\pi }{3}

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]

The branch with range \left [ 0,\pi \right ] is called the principal value branch of the function \cos^{-1} and domain of the function \cos^{-1} is \left [ -1, 1\right ].

Thus, \cos ^{-1}:[-1,1] \rightarrow[0, \pi]

Given :

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

Explanation :

Let is first solve for  \cos ^{-1} \frac{\sqrt{3}}{2}.

Let,  y=\cos ^{-1} \frac{\sqrt{3}}{2}                                                         .....(i)

       \begin{aligned} &\cos y=\frac{\sqrt{3}}{2} \\ &\cos y=\cos \frac{\pi}{6} \\ &y=\frac{\pi}{6} \\ &\cos ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{6} \end{aligned}

\because The range of principal value branch of \cos^{-1} is \left [ 0,\pi \right ].

\therefore \; \; \; \; \; \; \; \; \quad \cos ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{6} \in[0, \pi]

The principal value of  \cos ^{-1} \frac{\sqrt{3}}{2} is \frac{\pi}{6}                                                                         .....(ii)

Now,

       \begin{aligned} &\tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\} \\ &=\tan ^{-1}\left\{2 \sin \left(4 \times \frac{\pi}{6}\right)\right\} \end{aligned}                                                             [From equation (ii)]

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

      =\tan ^{-1}\left(2 \times \frac{\sqrt{3}}{2}\right)                                                                                  \left[\because \sin \frac{2 \pi}{3}=\frac{\sqrt{3}}{2}\right]

     =\tan ^{-1}(\sqrt{3})                                                                                                   .....(iii)

  Now, 

Let     x=\tan ^{-1}(\sqrt{3})                                                                                          ......(iv)

          \begin{aligned} &\tan x=\sqrt{3} \\ &\tan x=\tan \frac{\pi}{3} \\ &x=\frac{\pi}{3} \end{aligned}         

          \tan ^{-1}(\sqrt{3})=\frac{\pi}{3}                                                                                      [From equation iv]

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

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

\therefore    Principal value of {\tan ^{-1}}(\sqrt{3}) is \frac{\pi}{3}.

Hence principal vaue of  \tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}is \frac{\pi}{3}.   

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