# [SOLVED] Help with simple Trig Identity Q

• May 12th 2009, 06:56 PM
demo1
[SOLVED] Help with simple Trig Identity Q
sinx + tanx
__________ = tanx
cosx + 1

Thanks!
• May 12th 2009, 07:14 PM
TwistedOne151
$\displaystyle \frac{\sin{x}+\tan{x}}{\cos{x}+1}=\frac{\sin{x}\fr ac{\cos{x}}{\cos{x}}+\tan{x}}{\cos{x}+1}$
$\displaystyle =\frac{\cos{x}\frac{\sin{x}}{\cos{x}}+\tan{x}}{\co s{x}+1}$
$\displaystyle =\frac{\cos{x}\tan{x}+\tan{x}}{\cos{x}+1}$
you should be able to continue from here.

--Kevin C.
• May 12th 2009, 07:16 PM
skeeter
Quote:

Originally Posted by demo1
sinx + tanx
__________ = tanx
cosx + 1

Thanks!

manipulating the numerator on the left side ...

$\displaystyle \sin{x} + \tan{x}$

$\displaystyle \frac{\sin{x}\cos{x}}{\cos{x}} + \frac{\sin{x}}{\cos{x}}$

$\displaystyle \frac{\sin{x}\cos{x} + \sin{x}}{\cos{x}}$

$\displaystyle \frac{\sin{x}(\cos{x} + 1)}{\cos{x}}$

now divide the last expression by the denominator in the original fraction
• May 12th 2009, 07:31 PM
demo1
thanks!
• May 12th 2009, 07:33 PM
demo1
Solved