Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Prove that

    11. \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right ) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x,\;\;-\frac{1}{\sqrt2}\leq x\leq 1

            [Hint: Put x = \cos 2\theta]

Answers (1)

best_answer

By using the Hint we will put x = \cos 2\theta;

we get then,

=\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right )

=\tan^{-1}\left(\frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2 \theta} + \sqrt{2\sin^2\theta}} \right )

=\tan^{-1}\left(\frac{\sqrt2{\cos \theta} - \sqrt2{\sin\theta}}{\sqrt2{\cos \theta} + \sqrt2{\sin\theta}} \right )

=\tan^{-1}\left(\frac{{\cos \theta} - {\sin\theta}}{{\cos \theta} + {\sin\theta}} \right )   dividing numerator and denominator by \cos \theta,

we get,

= \tan^{-1}\left ( \frac{1-\tan \theta}{1+\tan \theta} \right )

= \tan^{-1} 1 - \tan^{-1} (\tan \theta)      using the formula  \left [ \tan^{-1}x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy} \right ]

= \frac{\pi}{4} - \theta = \frac{\pi}{4}- \frac{1}{2}\cos^{-1}x  

As L.H.S = R.H.S

Hence proved

 

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 Supplementary Exam 2025: Registrations open for Class 10, 12 private students today
anam-khan

CBSE opens application to access 10th answer scripts; verification, revaluation for Class 12 starts on May 31
anam-khan

CBSE supplementary exams 2025 from July 15; form submission starts on May 30

Latest Question