For the principal values, evaluate the following:

Question:

For the principal values, evaluate the following:

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

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

(iii) $\sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}(-2)\right\}\right]$

(iv) $\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)$

Solution:

(i)

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

$=-\sin ^{-1}\left(\sin \frac{\pi}{3}\right)+\operatorname{cosec}^{-1}\left[\operatorname{cosec}\left(-\frac{\pi}{3}\right)\right]$

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

$=-\frac{2 \pi}{3}$

(ii)

$\sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})=\sec ^{-1}\left(\sec \frac{\pi}{4}\right)+2 \operatorname{cosec}^{-1}\left[\operatorname{cosec}\left(-\frac{\pi}{4}\right)\right]$

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

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

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

(iii)

$\sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}(-2)\right\}\right]=\sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}\left(\operatorname{cosec}-\frac{\pi}{6}\right)\right\}\right]$

$=\sin ^{-1}\left[\cos \left\{-\frac{\pi}{3}\right\}\right]$

$=\sin ^{-1}\left[\cos \left(\frac{\pi}{3}\right)\right]$

$=\sin ^{-1}\left(\frac{1}{2}\right)$

$=\sin ^{-1}\left(\sin \frac{\pi}{6}\right)$

$=\frac{\pi}{6}$

(iv)

$\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)=\operatorname{cosec}^{-1}\left[2 \times\left(-\frac{1}{\sqrt{3}}\right)\right]$

$=\operatorname{cosec}^{-1}\left[-\frac{2}{\sqrt{3}}\right]$

$=\operatorname{cosec}^{-1}\left[\operatorname{cosec}\left(-\frac{\pi}{3}\right)\right]$

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

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now