1. ## Proving trig identities

Hi, I've been working on a question for ages and can't find an answer. This is the first proof I've been asked to do also.

prove:

$\frac{\sin(x)}{1-\cos(x)} = \csc(x) + \cot(x)$

Am I right in to work the LHS like this:

$\sin(x) = \sqrt{1-\cos^2(x)}$

So,

$\sqrt{1-\cos^2(x)} = 1-\cos(x)$ ..? For $\frac{\sin(x)}{\sin(x)}$

So LHS =1?
And what do do with the RHS?

2. ## Re: Proving trig identities

Hint :

Multiply both numerator and denominator of the LHS fraction by :

$1+\cos x$

3. ## Re: Proving trig identities

Bigger hint?

So,

$\frac{\sin(\Theta)}{1-cos(\Theta)}\times\frac{1+cos(\Theta)}{1+cos(\Thet a)}=\frac{sin(\Theta)\times(1+\cos(\Theta)}{1-\cos^2(\Theta)}=\frac{1+\cos(\Theta)}{1-\cos(\Theta)}$...

4. ## Re: Proving trig identities

Originally Posted by camjerlams
Bigger hint?

So,

$\frac{\sin(\Theta)}{1-cos(\Theta)}\times\frac{1+cos(\Theta)}{1+cos(\Thet a)}=\frac{sin(\Theta)\times(1+\cos(\Theta)}{1-\cos^2(\Theta)}=\frac{1+\cos(\Theta)}{1-\cos(\Theta)}$...
Actually

\displaystyle \begin{align*} \frac{\sin{(\theta)} \left[ 1 + \cos{(\theta)} \right] }{1 - \cos^2{(\theta)}} &\equiv \frac{\sin{(\theta)} \left[ 1 + \cos{(\theta)} \right] }{ \sin^2{(\theta)} } \\ &\equiv \frac{1 + \cos{(\theta)}}{\sin{(\theta)}} \\ &\equiv \frac{1}{\sin{(\theta)}} + \frac{\cos{(\theta)}}{\sin{(\theta)}} \\ &\equiv \csc{(\theta)} + \cot{(\theta)} \end{align*}

5. ## Re: Proving trig identities

Ah yeah I totally didn't see the sin^2(x), thanks for that