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  • Need solution for RD Sharma maths Class 12 Chapter Inverse Trignometric Function Exercise 3.3 Question 3 Sub question (ii) maths text book solution.

Need solution for RD Sharma maths Class 12 Chapter Inverse Trignometric Function Exercise 3.3 Question 3 Sub question (ii) maths text book solution.

Answers (1)

Answer : \frac{-3\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}.

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

Explanation :

Let us first solve for \tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)

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

        \begin{aligned} &\tan x=\frac{-1}{\sqrt{3}} \\ &\tan x=-\tan \frac{\pi}{6} \end{aligned}                                                                                    \left[\because \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]

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

      \begin{aligned} &x=\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 \; \; \; \; \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}                                                      ... (ii)

Now,

Let us solve for   \tan ^{-1}(-\sqrt{3})

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

       \begin{aligned} &\tan y=-\sqrt{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right] \\ &\tan y=-\tan \left(\frac{\pi}{3}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \tan (-\theta)=-\tan \theta] \end{aligned}

       \begin{aligned} &\tan y=\tan \left(\frac{-\pi}{3}\right) \\ &y=\frac{-\pi}{3} \end{aligned}                                                                        [From equation (iii)]

       \tan ^{-1}(-\sqrt{3})=\frac{-\pi}{3}

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

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

Hence principal value of \tan ^{-1}(-\sqrt{3})=\frac{-\pi}{3}                                                            ...(iv)

Now,

Let us solve for  \tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)

                \tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)=\tan ^{-1}(-1) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \sin \frac{\pi}{2}=1\right]

               \begin{aligned} &=\tan ^{-1}\left(\tan \left(\frac{-\pi}{4}\right)\right) \\ &=\frac{-\pi}{4} \\ \tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right) &=\frac{-\pi}{4} \end{aligned}

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

 \therefore \tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)=\frac{-\pi}{4} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Hence principal value \tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)=\frac{-\pi}{4}                                                          .....(v)

Now consider

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

From equation (ii), (iv) and (v)

\begin{aligned} &=-\frac{\pi}{6}+\left(-\frac{\pi}{3}\right)+\left(\frac{-\pi}{4}\right) \\ &=-\frac{\pi}{6}-\frac{\pi}{3}-\frac{\pi}{4} \\ &=\frac{-2 \pi-4 \pi-3 \pi}{12} \\ &=\frac{-9 \pi}{12} \\ &=\frac{-3 \pi}{4} \end{aligned}                                                   

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