# Prove this identity

• Nov 21st 2010, 04:19 PM
jellyksong
Prove this identity
I'm not sure how to prove these expressions true...the question is:

Prove that each expression is an identity for all values for which it is defined.

$\displaystyle \frac {sin\theta} {sin\theta+cos\theta} = \frac {tan\theta} {1+tan\theta}$

and,

$\displaystyle \frac {sin^2\theta+2cos\theta-1} {sin^2+3cos\theta-3} = \frac {1} {1-sec\theta}$

Thank you! :D
• Nov 21st 2010, 04:24 PM
harish21
$\displaystyle \dfrac{\sin x}{\sin x+\cos x}$

divide the numerator and denominator by cosx

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

For the other question, show some effort. Where exactly are you stuck?
• Nov 21st 2010, 04:26 PM
jellyksong
Oh, wow, that was so obvious hahaha
Thanks!! :D

Any ideas on the second one?
• Nov 21st 2010, 04:43 PM
harish21
$\displaystyle \dfrac{\sin^2 x+2\cos x-1}{\sin^2 x+3\cos x-3}$

$\displaystyle = \dfrac{1-\cos^2 x+2\cos x-1}{1-\cos^2 x+3\cos x-3}$

$\displaystyle = \dfrac{-\cos^2 x+2\cos x}{-\cos^2 x+3\cos x-2}$

$\displaystyle =\dfrac{-\cos x(cos x-2)}{-(\cos^2 x-3\cos x+2)}$

$\displaystyle =\dfrac{\cos x(cos x-2)}{(\cos^2 x-3\cos x+2)}$

factorize the denominator and change all cosx into secx..
• Nov 21st 2010, 04:56 PM
jellyksong
That makes sense now, thank you! :D