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Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Functions exercise 3.5 question 3 sub question (ii)

Answers (1)


The principal value branch of the function \sec ^{-1} \text { is }[0, \pi]-\left[\frac{-\pi}{2}\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 between these intervals

Given: \sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})

Explanation: \sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})                                                                                  ….(i)

Let us first solve for \sec ^{-1}(\sqrt{2})

Let x=\sec ^{-1}(\sqrt{2})

\sec x= (\sqrt{2})

\sec x=\sec \frac{\pi}{4}                                                                                                                              \left [\because \sec \frac{\pi}{4}= \sqrt{2} \right ]

x = \frac{\pi}{4}

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

The principal value of \sec ^{-1}(\sqrt{2}) is \frac{\pi}{4}                                                                                                        …. (ii)

Now let us find 2 \operatorname{cosec}^{-1}(-\sqrt{2})

Let y= 2 \operatorname{cosec}^{-1}(-\sqrt{2})

\operatorname{cosec} y= 2 x (\sqrt{2})

\operatorname{cosec} y= 2 \operatorname{cosec} \left ( \frac{\pi}{4} \right )

y = 2\times\frac{-\pi }{4}

y = \frac{-\pi }{2}

\therefore 2 \operatorname{cosec}^{-1}(-\sqrt{2})=\frac{-\pi}{2} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}

The principal value of 2 \operatorname{cosec}^{-1}(-\sqrt{2}) is-\frac{-\pi}{2}                                                                           …. (iii)

Now from equation (i)

\sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})

Putting the values of \sec ^{-1}(\sqrt{2}) \: \text{and }2 \operatorname{cosec}^{-1}(-\sqrt{2})                                                      [from (ii) and (iii)]

\therefore \frac{\pi}{4}+\left ( \frac{-\pi }{2} \right )

\frac{\pi}{4}- \frac{\pi }{2}



Therefore, the principal value of \sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})  is \frac{-\pi}{4}

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