Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 43 math

Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 43 math

Answers (1)

best_answer

Answer:  \frac{\pi}{3}

Given:\tan ^{-1}\left[2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right]

Hint: Try to solve \cos ^{-1}  and then \sin function.

Solution:

\begin{aligned} \tan ^{-1}\left[2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right] &=\tan ^{-1}\left\{2 \sin \left[\cos ^{-1} 2\left(\frac{\sqrt{3}}{2}\right)^{2}-1\right]\right.\\ &=\tan ^{-1}\left[2 \sin \left(\cos ^{-1} \frac{1}{2}\right)\right] = \frac{\pi}{3}\end{aligned}

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 Class 10, 12 re-evaluation, verification LOT 3 results 2023 declared at cbseresults.nic.in
anam-khan

CBSE asks schools to keep LOC, registration data of students ready for board exams 2024
anam-khan

CBSE 12th supplementary result verification application from tomorrow, board announces revaluation details

Latest Question