# difficult question!!!!

• Jul 17th 2009, 08:07 AM
jashansinghal
difficult question!!!!
$\displaystyle \log_{\tan(\frac{\pi}{4}+\frac{x}{2})} (\sec x + \tan x)-\log_{\tan(\frac{\pi}{4}-\frac{x}{2})} (\sec x - \tan x)$
• Jul 17th 2009, 10:05 AM
Hello jashansinghal
Quote:

Originally Posted by jashansinghal
$\displaystyle \log_{\tan(\frac{\pi}{4}+\frac{x}{2})} (\sec x + \tan x)-\log_{\tan(\frac{\pi}{4}-\frac{x}{2})} (\sec x - \tan x)$

So what is the question?

$\displaystyle \sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}=\frac{1+\sin x}{\cos x}=\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin \frac{x}{2}\cos\frac{x}{2}}{\cos x}=$
$\displaystyle =\frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right )^2}{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}+\sin\f rac{x}{2}\right)}=$
$\displaystyle =\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{ x}{2}-\sin\frac{x}{2}}=\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}=\tan\left(\frac{\pi}{4}+\frac{x}{ 2}\right)$
Then $\displaystyle \log_{\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)}( \sec x+\tan x)=1$
In the same way, $\displaystyle \log_{\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}(\sec x-\tan x)=1$