logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
0 votes

Find the value of $\tan\big[\large\frac{1}{2}$$\big(\sin^{-1}\large\frac{2x}{1+x^2}+$$\cos^{-1}\large\frac{1-y^2}{1+y^2}\big)\big],$$\mid x\mid < 1,y > 0$ and $xy <1$

$\begin{array}{1 1}(a)\;\large\frac{xy}{1-xy}&(b)\;\large\frac{x-y}{1-xy}\\(c)\;\large\frac{x+y}{1-xy}&(d)\;None\;of\;these\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
We know that
$\sin^{-1}\large\frac{2x}{1+x^2}$$=2\tan^{-1}x$
$\cos^{-1}\large\frac{1-y^2}{1+y^2}$$=2\tan^{-1}y$
$\tan\big[\large\frac{1}{2}$$\big(\tan^{-1}x+\tan^{-1}y\big)\big]=\tan\big[\tan^{-1}x+\tan^{-1}y\big]$
$\Rightarrow \tan\big[\tan^{-1}\big(\large\frac{x+y}{1-xy}\big)\big]$
$\tan(\tan^{-1}x)=x$
$\Rightarrow \large\frac{x+y}{1-xy}$
Hence (c) is the correct answer.
answered Oct 17, 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
...