Prove this identity

**jellyksong** · November 21st 2010, 04:19 PM

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.

$<br /> \frac {sin\theta} {sin\theta+cos\theta} = \frac {tan\theta} {1+tan\theta}<br />$

and,

$<br /> \frac {sin^2\theta+2cos\theta-1} {sin^2+3cos\theta-3} = \frac {1} {1-sec\theta}<br />$

Thank you!

**harish21** · November 21st 2010, 04:24 PM

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

divide the numerator and denominator by cosx

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

For the other question, show some effort. Where exactly are you stuck?

**jellyksong** · November 21st 2010, 04:26 PM

Oh, wow, that was so obvious hahaha
Thanks!!

Any ideas on the second one?

**harish21** · November 21st 2010, 04:43 PM

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

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

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

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

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

factorize the denominator and change all cosx into secx..

**jellyksong** · November 21st 2010, 04:56 PM

That makes sense now, thank you!

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