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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)

$\frac{-\pi}{4}$

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)]

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

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

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

$\frac{-\pi}{4}$

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

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