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 R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 7 Sub Question 1 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 7 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer: \frac{\pi}{4}

Given: Given that  \tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}

Hint: First we solve 2 \sin ^{-1} \frac{1}{2} Since \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}

Solution: We have,

\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}

=\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}                                                            \left[\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\right]

\begin{aligned} &=\tan ^{-1}\left\{2 \cos \frac{\pi}{3}\right\} \\ &=\tan ^{-1}\left\{2 \times \frac{1}{2}\right\} \end{aligned}                                                                        \left[\cos \frac{\pi}{3}=\frac{1}{2}\right]

\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}                                                                                       \left[\tan ^{-1}(1)=\frac{\pi}{4}\right]

This is the required solution.

Posted by

infoexpert21

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 Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4
anam-khan

When will CBSE Board exam 2025 date sheet be released for Class 10, 12?

Latest Question