1. ## Proving the identity

Prove idenity: 1 + sin theta / cos theta = cos theta / 1 - sin theta

I cant figure out which idenity!?

Show steps if at all possible thanks in advance.

2. Hi

$\displaystyle cos^2\theta + sin^2\theta = 1$

$\displaystyle cos^2\theta = 1 - sin^2\theta$

$\displaystyle cos \theta cos \theta = (1 - sin\theta) (1 + sin\theta)$

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

3. ## ?

Still a bit confused, The more i think about it, Maybe I just dont understand the goal of proving an idenity, I thought it was to get the problem into the form of one of the basic idenitys such as tan = sin/cos.

4. Hello, Garface!

You're expected to know this identity:

. . $\displaystyle \sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \begin{Bmatrix}\sin^2\!\theta \:=\:1 - \cos^2\!\theta \\ \cos^2\!\theta \:=\:1-\sin^2\!\theta \end{Bmatrix}$

Prove the idenity: .$\displaystyle \frac{1 + \sin\theta}{\cos\theta} \:=\:\frac{\cos\theta}{1 - \sin\theta }$

Multiply the right side by $\displaystyle \frac{1+\sin\theta}{1+\sin\theta}$

. . $\displaystyle \frac{\cos\theta}{1-\sin\theta}\cdot\frac{1+\sin\theta}{1 + \sin\theta} \;=\;\frac{\cos\theta(1+\sin\theta)}{1-\sin^2\!\theta} \;=\;\frac{\cos(1+\sin\theta)}{\cos^2\!\theta} \;=\; \frac{1+\sin\theta}{\cos\theta}$