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

If $ y=tan^{-1}\left \{\large\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right \} $,find $\large\frac{dy}{dx}.$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $1+\cos 2\theta=2\cos^2\theta$
  • $1-\cos 2\theta=2\sin^2\theta$
  • $\tan\large\frac{\pi}{4}$$=1$
Step 1:
$ y=\tan^{-1}\left \{\large\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right \} $
Put $x^2=\cos 2\theta$
$\therefore y=tan^{-1}\bigg(\large\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\bigg)$
$1+\cos 2\theta=2\cos^2\theta$
$1-\cos 2\theta=2\sin^2\theta$
$\therefore y=\tan^{-1}\bigg(\large\frac{\sqrt 2(\cos\theta-\sin\theta)}{\sqrt 2(\cos\theta+\sin\theta)}\bigg)$
Step 2:
Dividing by $\cos\theta$ we get,
$y=\tan^{-1}\bigg(\large\frac{1-\tan \theta}{1+\tan\theta}\bigg)$
$\tan\large\frac{\pi}{4}$$=1$
$\;\;=\tan^{-1}\bigg(\large\frac{\tan \pi/4-\tan\theta}{\tan\pi/4+\tan\theta}\bigg)$
This is of the form $\tan(A-B)$
$\;\;\;=\tan^{-1}\big(\tan(\large\frac{\pi}{4}$$-\theta)\big)$
Step 3:
$y=\large\frac{\pi}{4}$$-\theta$
But $\theta=\cos^{-1}(x^2)$
$\Rightarrow y=\large\frac{\pi}{4}$$-\cos^{-1}(x^2)$
$\therefore \large\frac{dy}{dx}$$=0-\big(-\large\frac{1}{2}.\frac{2x}{\sqrt{1-x^4}}\big)$
$\Rightarrow \large\frac{x}{\sqrt{1-x^4}}$
answered Oct 3, 2013 by sreemathi.v
 

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
...