logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Solve the following equation for x : $ \tan^{-1} \bigg( \large\frac{1-x}{1+x} \bigg) = \large\frac{1}{2}\tan^{-1}(x), x > 0 $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • \(tan(x-y)=\frac{tanx-tany}{1+tanx.tany}\)
  • \(tan\frac{\pi}{4}=1\)
Let \(x=tan\theta,\:\Rightarrow\:\theta=tan^{-1}x\)
Substituting the value of x in the given equation, we have
\(tan^{-1}\bigg(\frac{1-tan\theta}{1+tan\theta}\bigg)=\frac{1}{2}tan^{-1}tan\theta\)
We know that \(tan\frac{\pi}{4}=1\)
\(\Rightarrow\:\large\:tan^{-1}\bigg(\large\frac{tan\frac{\pi}{4}-tan\theta}{1+tan\frac{\pi}{4}.tan\theta}\bigg)=\large\frac{1}{2}.\theta\)
\(\Rightarrow\:\large\:tan^{-1}tan(\frac{\pi}{4}-\theta)=\large\:\frac{1}{2}\theta\)
\(\Rightarrow\:\large\frac{\pi}{4}-\theta=\large\frac{\theta}{2}\)
\(\Rightarrow\:\large\frac{3\theta}{2}=\large\frac{\pi}{4}\)
\(\Rightarrow\:\large\theta=\large\frac{\pi}{6}=\large\:tan^{-1}x\)
\(\Rightarrow\:\large\:x=\large\:tan\frac{\pi}{6}=\large\frac{1}{\sqrt3}\)
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 21, 2013 by rvidyagovindarajan_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...