logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Prove that $tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac{1}{2}cos^{-1}x^2$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • \( put \: x^2 = cos\theta\)
  • \( \Rightarrow \theta = cos^{-1}x^2\)
  • \( 1+cos\theta = 2cos^2\large\frac{\theta}{2} \: \: ; \: \; 1-cos\theta=2sin^2\large\frac{\theta}{2}\)
  • \(\large\frac{tanA+tanB}{1-tanAtanB}=tan(A+B)\)
  • \(tan\large\frac{\pi}{4}=1\)
L.H.S.
put \(x^2=cos\theta\)
\( \Rightarrow \theta = cos^{-1}x^2\)
\(\sqrt{1+x^2}=\sqrt{1+cos\theta}=\sqrt{2cos^2\large\frac{\theta}{2}}=\sqrt{2}cos\large\frac{\theta}{2}\)
\(\sqrt{1-x^2}=\sqrt{1-cos\theta}=\sqrt{2sin^2\large\frac{\theta}{2}}=\sqrt{2}sin\large\frac{\theta}{2}\)
\(\large\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\large\frac{\sqrt{2}cos\large\frac{\theta}{2}+\sqrt{2}sin\large\frac{\theta}{2}}{\sqrt{2}cos\large\frac{\theta}{2}-\sqrt{2}sin\large\frac{\theta}{2}}\)
\(tan^{-1}\large\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=tan^{-1} \bigg[ \large\frac{\sqrt{1+cos\theta}+\sqrt{1-cos\theta}}{\sqrt{1+cos\theta}-\sqrt{1-cos\theta}}\bigg] \)
\( = tan^{-1} \bigg[ \large\frac{cos\frac{\theta}{2}+sin\large\frac{\theta}{2}}{cos\large\frac{\theta}{2}-sin\large\frac{\theta}{2}} \bigg]\)
 
Divide num.and den. by \(cos\large\frac{\theta}{2}\)
\(=tan^{-1}\bigg[\large\frac{1+\large\frac{sin\large\frac{\theta}{2}}{cos\large\frac{\theta}{2}}}{1-\large\frac{sin\frac{\theta}{2}}{cos\large\frac{\theta}{2}}}\bigg]\)=\(tan^{-1}\big(\frac{1+tan\large\frac{\theta}{2}}{1-tan\large\frac{\theta}{2}}\big)\)
 
By writing \(1=tan\large\frac{\pi}{4}\) in the numerator we get
\(tan^{-1}\bigg(\frac{tan\large\frac{\pi}{4}+tan\large\frac{\theta}{2}}{1-tan\large\frac{\pi}{4}.tan\large\frac{\theta}{4}}\bigg)\)
By taking A =\(\large\frac{\pi}{4}\:and\:B=\large\frac{\theta}{2}\) in the formula of tan(A+B). we get
\(\large\frac{tan\large\frac{\pi}{4}+tan\large\frac{\theta}{2}}{1-tan\large\frac{\pi}{4}.tan\large\frac{\theta}{4}}=tan(\large\frac{\pi}{4}+\large\frac{\theta}{2})\)
\(\Rightarrow\: tan^{-1}tan \bigg( \large\frac{\pi}{4}+\large\frac{\theta}{2} \bigg) = \large\frac{\pi}{4}+\large\frac{\theta}{2}\)
 
By substituting the value of \(\theta=cos^{-1}x\) \(\Rightarrow\:\large\frac{\pi}{4}+\large\frac{1}{2} cos^{-1}x \)
=R.H.S

 

answered Mar 2, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 
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
...