Ask Questions, Get Answers

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

Evaluate $\tan^{-1} \left (\large \frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{{\sqrt {1+x}}+{\sqrt {1-x}}} \right )$$,\frac{-1}{\sqrt2} \leq x \leq 1$

Can you answer this question?

1 Answer

0 votes
  • $ 1+\cos\theta = 2 \cos^2\frac{\theta}{2} $
  • $ 1-\cos\theta=2 \sin^2\frac{\theta}{2}$
  • $\tan(A-B) = \large \frac{\tan A-\tan B}{1+\tan A \tan B}$
  • \(\tan\frac{\pi}{4}=1\)
Given $\tan^{-1} \left ( \large \frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{{\sqrt {1+x}}+{\sqrt {1-x}}} \right ),-\frac{1}{\sqrt2} \leq x \leq 1$
Let \( x=\cos\theta\) \( \Rightarrow \theta = \cos^{-1}x\)
$\rightarrow $\(\sqrt{1+x}=\sqrt{1+\cos\theta}=\sqrt{2\cos^2\frac{\theta}{2}}=\sqrt{2}\cos\frac{\theta}{2}\)
$\rightarrow$ \(\sqrt{1-x}=\sqrt{1-\cos\theta}=\sqrt{2\sin^2\frac{\theta}{2}}=\sqrt{2}\sin\frac{\theta}{2}\)
$\rightarrow$ \( \sqrt{1+x}-\sqrt{1-x} \)\(=\sqrt{2\cos\frac{\theta}{2}}-\sqrt{2\sin\frac{\theta}{2}} =\) \( \sqrt 2\;(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}) \)
$\rightarrow$ \( \sqrt{1+x}+\sqrt{1-x} \)\(=\sqrt{2\cos\frac{\theta}{2}}+\sqrt{2\sin\frac{\theta}{2}} \) \(= \sqrt 2\;(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}) \)
$\Rightarrow $ \(\large \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}= \large\frac{ \sqrt 2\;(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})}{ \sqrt 2\;(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})}\)
Dividing numerator and denonimator by \(\sqrt 2 \) and \(\cos\frac{\theta}{2}\), this reduces to:
$\Rightarrow $\(\;\large \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \) = \(\Large \frac{1-\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}}{1+\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}}\) = \(\large \frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}\)
Since \(\tan\frac{\pi}{4} = 1\) we can rewrite this as \( \large \frac{\tan\frac{\pi}{4}-\tan\frac{\theta}{2}}{1+\tan\frac{\pi}{4}.\tan\frac{\theta}{4}}\)
We know that $\tan(A-B) = \large \frac{\tan A-\tan B}{1+\tan A \tan B}$
By substituting for $A =$\(\large \frac{\pi}{4}\), \(\:B=\large\frac{\theta}{2}\), we get \(\large \frac{\tan\frac{\pi}{4}-\tan\frac{\theta}{2}}{1+\tan\frac{\pi}{4}.\tan\frac{\theta}{4}}\) \(=\tan(\frac{\pi}{4}-\frac{\theta}{2})\)
\(\Rightarrow\: \tan^{-1} \tan \big( \large \frac{\pi}{4}-\frac{\theta}{2}\)\( \big) \) \(= \large \frac{\pi}{4}-\frac{\theta}{2}\)
By substituting the value of \(\theta=\cos^{-1}x\) \(\Rightarrow\:\large \frac{\pi}{4}-\frac{1}{2}\)\(\cos^{-1}x \)
Answer: $ \tan^{-1} \left ( \large \frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{{\sqrt {1+x}}+{\sqrt {1-x}}} \right ),-\frac{1}{\sqrt2} \leq x \leq 1 = $ \(\large \frac{\pi}{4}-\frac{1}{2}\)\(\cos^{-1}x \)
answered Mar 14, 2013 by balaji.thirumalai
edited Dec 19, 2013 by balaji.thirumalai

Related questions

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