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  • Provide solution for RD Sharma maths Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 2 Subquestion (i) maths textbook solution.

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

Answers (1)

Answer : \frac{\pi }{2}

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} \text { is }[-1,1].

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

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

Explanation :

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

Let  y=\tan ^{-1}(-1)                                                                              ....(i)

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

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

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

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

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

\begin{aligned} &\tan ^{-1}(-1)=\frac{-\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\\ &\therefore \text { Principal value of }\\ &\tan ^{-1}(-1)=\frac{-\pi}{4} \end{aligned}                                                                            .....(ii)

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

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

\cos x=\left(\frac{-1}{\sqrt{2}}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]

\cos x=-\cos \frac{\pi}{4}

\cos x=\cos \left(\pi-\frac{\pi}{4}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \cos (\pi-\theta)=-\cos \theta]

\begin{aligned} &\cos x=\cos \left(\frac{3 \pi}{4}\right) \\ &x=\frac{3 \pi}{4} \\ &\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4} \end{aligned}                                                                            [From equation (iii)]

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

\therefore \quad \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4} \in[0, \pi]

\therefore \text { Principal value of } \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right) \text { is } \frac{3 \pi}{4} \text { . }                                                  .....(iv)

Now,

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

\begin{aligned} &=\frac{2 \pi}{4} \\ &=\frac{\pi}{2} \end{aligned}.

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