Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Solve the following equations: 14.

Solve the following equations:

    14. \tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x,\;(x>0)

Answers (1)

best_answer

Given equation is

\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x:

L.H.S can be written as;

\tan^{-1} \frac{1-x}{1+x} = \tan^{-1}1 - \tan^{-1}x 

Using the formula \left [ \tan^{-1}x -\tan^{-1}y = \tan^{-1} \frac{x-y}{1+xy} \right ]

So, we have \tan^{-1}1 - \tan^{-1}x = \frac{1}{2} \tan^{-1}x

\Rightarrow \tan^{-1}1= \frac{3}{2} \tan^{-1}x

\Rightarrow\frac{\pi}{4}= \frac{3}{2} \tan^{-1}x     

\Rightarrow \tan^{-1}x = \frac{\pi}{6}

 \Rightarrow x= \tan \frac{\pi}{6} = \frac{1}{\sqrt3}

Hence the value of x= \frac{1}{\sqrt3}.

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

'Award grace marks': CBSE's July 28 supplementary exam sparks student outrage over college admissions
anam-khan

CBSE Class 12 supplementary exam 2026 on July 28; LOC submission starts
anam-khan

CBSE OSM Evaluation Row: Class 12 revised marksheet sparks fresh debate, Vedant disputes board's explanation

Latest Question