Proving the trig identity $\frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$ without cross-multiplying I need to prove the following identity.

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

I want to prove it by deduction rather than cross multiplying.
 A: Here's a trigonograph:


$$\triangle QSR \sim \triangle RSP \quad\implies\quad \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta} \tag{$\star$}$$

You may recognize $\angle P = \theta/2$ (by the Inscribed Angle Theorem), in light of which, the ratios in $(\star)$ can be seen to have the common value $\tan(\theta/2)$.
A: $$\frac{1-\cos\theta}{\sin\theta}=\frac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\frac{1-\cos^2\theta}{\sin\theta(1+\cos\theta)}=\frac{\sin^2\theta}{\sin\theta(1+\cos\theta)}$$

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

Let $\displaystyle t=\tan\frac{\theta}{2}$
$$\frac{1-\cos\theta}{\sin\theta}=\frac{1-\frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}}=t$$
$$\frac{\sin\theta}{1+\cos\theta}=\frac{\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}}=t$$
A: 
$$\tan{\varphi}= \frac{1-\cos{\theta}}{\sin \theta} = \frac{\sin \theta}{1+\cos{\theta}}$$
A: HINT: $$1-\cos^2\theta=\sin^2\theta\implies\color{red}{(1-\cos\theta)}\color{blue}{(1+\cos\theta)}=\color{red}{\sin\theta}\color{blue}{\sin\theta}\implies\cdots$$
A: $$\frac{1-\cos\theta}{\sin\theta} = $$    $$   \frac{(1-\cos\theta) (1+\cos\theta)   }{\sin\theta     (1+\cos\theta)   } = $$
$$       \frac{\sin^2\theta}{\sin \theta(1+\cos\theta)}= $$
$$        \frac{\sin\theta}{1+\cos\theta}$$
A: I think your imprecations against "cross-multiplying" are silly, but does
this fit the bill for you?
$$\frac{1-\cos\theta}{\sin\theta}=\frac{1-(1-2\sin^2(\theta/2))}
{2\sin(\theta/2)\cos(\theta/2)}=\frac{\sin(\theta/2)}{\cos(\theta/2)}$$
and
$$\frac{\sin\theta}{1+\cos\theta}=\frac{2\sin(\theta/2)\cos(\theta/2)}
{(2\cos^2(\theta/2)-1)+1}=\frac{\sin(\theta/2)}{\cos(\theta/2)}.$$
But TheSimpliFire's proof is better!
A: I'm going to make a geometric proof for an angle $\theta \in (0,\pi/2)$. The argument can be modified for any angle but some segments take negative values in other quadrants.
First note that
$$\dfrac{1-\cos\theta}{\sin\theta} =  \dfrac{1}{\sin\theta} - \dfrac{\cos\theta}{\sin\theta} = \csc\theta - \cot\theta.$$
The geometrical definition of $\csc \theta$ correspongs with the segment $BF$ and the one for $\cot\theta$ corresponds with $CF$.
The ratio $\dfrac{\sin\theta}{1 + \cos\theta}$ is a bit more tricky. First note that $\sin\theta$ is $CJ$ and $\cos\theta$ is $BJ$. Since the circumference has radius $1$ then $DJ$ is $1 + \cos\theta$. By creating the line $EB$ paralel to $CD$ we construct a similar triangle to $CDJ$, namely $EBK$. That means that
$$\dfrac{\sin\theta}{1 + \cos\theta} = \dfrac{CJ}{DJ} = \dfrac{EK}{BK} = \tan \vartheta,$$
where $\vartheta$ is the angle $\angle EBK$. Let us find the value of $\vartheta$:
We see that the cord $CH$ is viewed from the centre $B$ with an angle of $2\theta$, that means that it is viewed from $D$ with an angle of $\theta$. Moreover, the line $DB$ bisects both this angles. That means that $\angle CDJ = \theta/2$ but $\angle CDJ = \angle EBK = \vartheta$, so
$$\dfrac{\sin\theta}{1 + \cos\theta} = \tan \dfrac{\theta}{2}$$
which is, geometrically, the segment $EG$.

Now we can state the problem in terms of those segments.
$$\csc\theta - \cot\theta = \tan \dfrac{\theta}{2}$$
becomes
$$BF - CF = EG \iff BF = CF + EG.$$
First of all, lets us observe the triangles $CBI$ and $EBG$. By the ASA rule, woth are congruent and $EG = CI$. So our problem becomes proving $BF = CF + CI$.
Since the points $F$, $C$ and $I$ are colinear, we can simply prove that $BF = FI$. Let us do that.

Note that $\pi/2 = \angle FBJ = \angle FBC + \angle CBJ = \angle FBC + \theta$. This means that $\angle FBC = \pi/2 - \theta$ and since $\angle BCF = \pi/2$ then $\angle BFC = \theta$.
Since $\angle CBI = \theta/2$ then $\angle FBI = \pi/2 - \theta/2$, but on the other hand, $\angle FIB = \angle CIB = \pi/2 - \theta/2$ which means that the triangle $FBI$ is isosceles with equal sides $BF = FI$ as desired. This completes the proof.


A: You could use the half angle formulae .
$\large\frac{1-\cos\theta}{\sin\theta}$
$=\large\frac{2\sin^2\frac\theta2}{2\sin\frac\theta2.cos\frac\theta2}$
$= \large\frac{2\sin\frac\theta2}{2\cos\frac\theta2}$ 
multiply both numerator and denominator by $\cos\frac\theta2$
$=\large\frac{ 2\sin\frac\theta2\cos\frac\theta2}{2\cos^2\frac\theta2}$
$= \large\frac{\sin\theta}{\cos\theta+1}$
