1. Help with Trigonometric Identities

I need to solve the trig identity: (x stands for theta)
1- (cos^2 x / 1 + sin x) = sin x

I get the first step
1 - ( 1 - sin ^ 2 x / 1 + sin x)
And i dont understand the 3rd step

1-(1-sin x)
Can someone please explain to me how the third step is gotten?
thanks

2. Originally Posted by mathmike
I need to solve the trig identity: (x stands for theta)
1- (cos^2 x / 1 + sin x) = sin x

I get the first step
1 - ( 1 - sin ^ 2 x / 1 + sin x)
And i dont understand the 3rd step
1-(1-sin x)
Can someone please explain to me how the third step is gotten?
thanks
yes, factor the numerator in your second step using the difference of two squares. that will cancel the denominator and give you what you want.

3. $\displaystyle \displaystyle {1 - \frac{cos^{2}x}{1+sinx} = 1 - \frac{1-sin^{2}x}{1+sinx} = 1 - \frac{(1+sinx)(1-sinx)}{1+sinx}= 1-(1-sinx) = sinx}$

note that:

$\displaystyle 1-sin^{2}x = (1)^2 -(sinx)^2 = (1+sinx)(1-sinx)$

because

$\displaystyle (a)^2-(b)^2 = (a+b)(a-b)$