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  • Explain solution RD Sharma class 12 chapter Inverse Trigonometric Functions exercise 3.5 question 3 sub question (i) maths

Explain solution RD Sharma class 12 chapter Inverse Trigonometric Functions exercise 3.5 question 3 sub question (i) maths

Answers (1)

\frac{-2\pi }{3}

Hints: The principal value branch of the function \sin ^{-1} is \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}

The principal value branch of the function \operatorname{cosec}^{-1} is \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}

Then find all the principal values b/w these intervals.

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

Explanation: \sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)                                                                                                …(i)

Let us solve for \sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)

Let x=\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)

\frac{-\sqrt{3}}{2}

\sin x = \sin \frac{\pi }{3}                                                                                                                        \left [\because \sin \frac{\pi}{3}= \frac{\sqrt{3}}{2} \right ]

   x = -\frac{\pi }{3}

\therefore \sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{-\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]                                                                                                                     … (ii)

Let us solve for \operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)

Let y=\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)

\operatorname{cosec} y =\left(\frac{-2}{\sqrt{3}}\right)

\operatorname{cosec} y = \operatorname{cosec} \frac{\pi}{3}                         \left [\because \operatorname{cosec} \frac{\pi}{3}= \frac{2}{\sqrt{3}} \right ]

y= \frac{-\pi}{3}

\therefore \operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\frac{-\pi}{3} \in\left[\frac{-\pi}{3}, \frac{\pi}{3}\right]                                                                                                                       ….(iii)

Now from equation (i)

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

Putting the values of \sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\text{ and }\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)                                                     [from (ii) (iii)]

Hence\left ( \frac{-\pi }{3} \right )+\left ( \frac{-\pi}{3} \right )

\frac{-\pi }{3} - \frac{-\pi}{3}

\frac{-2\pi }{3}

Therefore, the principal value of

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

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