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 for x : $ 2\tan^{-1}(\sin x)=\tan^{-1} (2\sec x),0 < x < \large\frac{\pi}{2} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • \( 2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}\)
By taking sinx in the place of x in the above formula of \(2tan^{-1}x\) we have
\( 2tan^{-1}sinx=tan^{-1} \bigg( \frac{2sinx}{1-sin^2x} \bigg)=tan^{-1} (2secx)\)
\( \Rightarrow\:\frac{2sinx}{cos^2x}=2secx\)
\(\large\frac{sinx}{cos^2x}=\large\frac{1}{cosx}\)
\(\Rightarrow\:\large\frac{sinx}{cosx}=1\:\Rightarrow\:\large\:tanx=1\)
\( \Rightarrow x=\frac{\pi}{4}\)
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 22, 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
...