Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 38

Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 38

Answers (1)

Answer:\frac{-\pi}{3}

Given:

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

Hint: \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)  is the range of the principal value branch of inverse sine function.

Solution:

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

Then,         

                 \begin{array}{c} \sin y=-\frac{\sqrt{3}}{2}=\sin \left(-\frac{\pi}{3}\right) \\\\ \end{array}

                \begin{array}{c} y=-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\\\ \end{array}

\begin{array}{c} \therefore \sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3} \end{array}

Posted by

infoexpert22

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Supplementary Result 2025 LIVE: Class 10, 12 compartment result soon on results.cbse.nic.in; updates
anam-khan

CBSE Supplementary Result 2025 LIVE: Class 10, 12 supplementary result soon on results.cbse.nic.in
anam-khan

CBSE Class 12 revaluation, verification results 2025 out at results.cbse.nic.in

Latest Question