Browse by Stream
QnA
Home
Search for Exam, Articles, Questions
QnA
get app
Go To Your Study Dashboard
  1. Home
  2. Find the values of each of the following: 11. tan inverse [2 cos (2sin inverse 1 / 2)]
# NCERT

Find the values of each of the following:

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

Answers (1)
D Divya Prakash Singh

Given equation:

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

So, solving the inner bracket first, we take the value of \sin x^{-1} \frac{1}{2} = x.

Then we have,

\sin x = \frac{1}{2} = \sin \left ( \frac{\pi}{6} \right )

Therefore, we can write \sin^{-1} \frac{1}{2} = \frac{\pi}{6}.

 \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 ]

= \tan^{-1}\left[2\cos\left(\frac{\pi}{3} \right ) \right ] = \tan^{-1}\left[2\times\left(\frac{1}{2} \right ) \right ] = \tan^{-1}1 = \frac{\pi}{4}.

 

Similar Questions

Related Chapters

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Buy Now
Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
Start Now
 
Exams
Articles
Questions