Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the values of the following: 11. tan inverse of 1 + cos inverse of - 1 / 2 plus sin inverse of - 1 / 2

Find the values of the following:

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

Answers (1)

best_answer

To find the values first we declare each term to some constant ;

tan^{-1}(1) = x , So we have \tan x = 1;

or \tan (\frac{\pi}{4}) = 1 

Therefore, x = \frac{\pi}{4} 

 cos^{-1}(\frac{-1}{2}) = y 

So, we have

 \cos y = \left ( \frac{-1}{2} \right ) = -\cos \left ( \frac{\pi}{3} \right ) = \cos(\pi - \frac{\pi}{3}) = \cos \left ( \frac{2\pi}{3} \right ).

Therefore y = \frac{2\pi}{3},

\sin^{-1}(\frac{-1}{2}) = z,

So we have;

\sin z = \frac{-1}{2}   or  -\sin (\frac{\pi}{6}) =\sin (\frac{-\pi}{6}) = \frac{-1}{2}

Therefore z = -\frac{\pi}{6}

Hence we can calculate the sum:

= \frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}

=\frac{3\pi + 8\pi -2\pi}{12} = \frac{9\pi}{12}=\frac{3\pi}{4}.

 

Posted by

Divya Prakash Singh

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