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

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

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}.

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

The branch with range \left[\frac{-\pi}{2}, \frac{\pi}{2}\right] is called the principal value branch.

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

Domain of functions \sin^{-1} and \cos^{-1} are \left [ -1,1 \right ] and \left [ -1,1 \right ] respectively.

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

Explanation :

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

Let

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

                     \tan y=\tan \frac{\pi}{4} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \frac{\pi}{4}=1\right]

                     \begin{aligned} &y=\frac{\pi}{4} \\ &\tan ^{-1}(1)=\frac{\pi}{4} \end{aligned}                                                                          [From equation (i)]

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

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

Therefore principal of value of \tan^{-1} (1) is  \frac{\pi}{4}                                                       ....(ii)

 Now,

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

Let  z=\sin ^{-1}\left(\frac{-1}{2}\right)                                                                                      ....(iii)

\begin{array}{ll} \sin z=\frac{-1}{2} &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\left[\because \sin \frac{\pi}{6}=\frac{1}{2}\right]} \\ \sin z=-\sin \frac{\pi}{6} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because \sin (-\theta)=-\sin \theta]} \end{array}

\begin{aligned} &\sin z=\sin \left(\frac{-\pi}{6}\right) \\ &z=\frac{-\pi}{6} \\ &\sin ^{-1}\left(\frac{-1}{2}\right)=\frac{-\pi}{6} \end{aligned}                                                                   [From equation (iii)]

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

\because \sin ^{-1}\left(\frac{-1}{2}\right)=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]

\therefore Principal value of \sin ^{-1}\left(\frac{-1}{2}\right) \text { is } \frac{-\pi}{6}                                                             ......(iv)

Now,

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

Let  x=\cos ^{-1}\left(\frac{-1}{2}\right)                                                                                            ....(v)

\begin{aligned} &\cos x=\frac{-1}{2} \\ &\cos x=-\cos \frac{\pi}{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned}

\cos x=\cos \left(\pi-\frac{\pi}{3}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \cos (\pi-\theta)=-\cos \theta]

\begin{aligned} &\cos x=\cos \left(\frac{2 \pi}{3}\right) \\ &x=\frac{2 \pi}{3} \\ &\cos ^{-1}\left(\frac{-1}{2}\right)=\frac{2 \pi}{3} \end{aligned}                                                               [From equation (v)]

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

\because \quad \cos ^{-1}\left(\frac{-1}{2}\right)=\frac{2 \pi}{3} \in[0, \pi]

\therefore Principal value of \cos ^{-1}\left(\frac{-1}{2}\right) is \frac{2\pi}{3}                                                                    ...(vi)

Now,

    \tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)

Putting the values from equations (ii), (iv) and (vi)

    \begin{aligned} &=\frac{\pi}{4}+\frac{2 \pi}{3}+\left(\frac{-\pi}{6}\right) \\ &=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \\ &=\frac{3 \pi+8 \pi-2 \pi}{12} \\ &=\frac{9 \pi}{12} \\ &=\frac{3 \pi}{4} \end{aligned}

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