Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$

Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and denominator by $1 + \sin A - \cos A$) has lead to a dead end.

Prove that, $$ \dfrac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \dfrac{A}{2} $$

Now that OP has understood how to prove this, here is a geometric proof for certain angles, just for fun :-)

Consider the figure:

$\displaystyle \triangle ABC$ is a right angled triangle with the right angle being at $\displaystyle C$.

$\displaystyle \angle{CAB} = A$ and $\displaystyle AB = 1$ and thus $\displaystyle BC = \sin A$ and $\displaystyle AC = \cos A$.

Now $\displaystyle AO$ is the angular bisector of $\displaystyle \angle{CAB}$. We select $\displaystyle O$ so that $\displaystyle O$ is the in-center (point of intersection of angular bisectors of a triangle). Let the in-radius $\displaystyle OD$ be $\displaystyle r$.

Now $\displaystyle BE = BF$ and $\displaystyle AE = AD$ and adding gives us $\displaystyle BF + AD = AB = 1$

Now $\displaystyle BF = BC - FC =BC - OD$ (as $\displaystyle ODCF$ is a square).

Thus $\displaystyle BF = \sin A - r$. Similarly $\displaystyle AD = \cos A - r$.

Thus $\displaystyle \sin A + \cos A - 2r = 1$.

Using $\displaystyle \triangle ADO$, $\displaystyle \tan \frac{A}{2} = \frac{OD}{AD} = \frac{r}{\cos A - r}$

Since $\displaystyle 2r = \sin A + \cos A -1 $ we get

$$\tan \frac{A}{2} = \frac{2r}{2\cos A - 2r} = \frac{\sin A + \cos A - 1}{\cos A - \sin A + 1}$$

It is easy to verify that

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

Incidently, the fact that in a right triangle (hypotenuse $c$), the in-radius is given by $ a + b -c = 2r$ and also by $r = \frac{\triangle}{s}$ ($\triangle$ is the area, $s$ is the semi-perimeter) can be used to prove the pythagoras theorem ($a^2 + b^2 = c^2$)!

$\textbf{Method 1.}$ You have

\begin{align*} \frac{1+ \sin{A} - \cos{A}}{1+\sin{A} + \cos{A}} &= \frac{ \cos^{2}\frac{A}{2} + \sin^{2}\frac{A}{2} + 2\cdot \sin\frac{A}{2}\cdot\cos\frac{A}{2} - \Bigl(\cos^{2}\frac{A}{2} - \sin^{2}\frac{A}{2}\Bigr)}{\cos^{2}\frac{A}{2} + \sin^{2}\frac{A}{2} + 2\cdot \sin\frac{A}{2}\cdot\cos\frac{A}{2} + \Bigl(\cos^{2}\frac{A}{2} -\sin^{2}\frac{A}{2}\Bigr)} \\ &=\frac{ \Bigl(\cos\frac{A}{2} + \sin\frac{A}{2}\Bigr)^{2} - \Bigl(\cos\frac{A}{2}+\sin\frac{A}{2}\Bigr) \cdot \Bigl(\cos\frac{A}{2} - \sin\frac{A}{2}\Bigr)}{\Bigl(\cos\frac{A}{2} + \sin\frac{A}{2}\Bigr)^{2} + \Bigl(\cos\frac{A}{2}+\sin\frac{A}{2}\Bigr) \cdot \Bigl(\cos\frac{A}{2} - \sin\frac{A}{2}\Bigr)} \\ &= \frac{ \cos\frac{A}{2} + \sin\frac{A}{2} - \cos\frac{A}{2} + \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2} + \cos\frac{A}{2} -\sin\frac{A}{2}} \qquad \Bigl[ \text{cancelling}\ \Bigl(\small\cos\frac{A}{2} +\sin\frac{A}{2}\Bigr) \ \Bigr] \\ &= \tan\frac{A}{2} \end{align*}

$\textbf{Method 2.}$ Here is another way of seeing this. By using something called Componendo and Dividendo.

• http://en.wikipedia.org/wiki/Componendo_and_dividendo

Let $$\frac{1+\sin{A} -\cos{A}}{1+\sin{A} + \cos{A}} = \frac{k}{1}$$ Now applying componendo and dividendo we get $$\frac{1+ \sin{A}}{-\cos{A}} = \frac{k+1}{k-1} \Longrightarrow \frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{-\cos\frac{A}{2}+\sin\frac{A}{2}} = \frac{k+1}{k-1} \qquad (1)$$

Again using componendo dividendo on $(1)$ we get $$\frac{\cos\frac{A}{2}+\sin\frac{A}{2} -\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2} +\cos\frac{A}{2}-\sin\frac{A}{2}} = \frac{k+1 +k-1}{k+1-k+1} \Longrightarrow k=\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}=\tan\frac{A}{2}$$

$\textbf{Method 3.}$ Another way of looking into this would be, \begin{align*} \frac{1+\sin{A}-\cos{A}}{1+\sin{A}+\cos{A}} =\frac{(1-\cos{A})+\sin{A}}{(1+\cos{A}) + \sin{A}} &= \frac{ 2\:\sin^{2}\frac{A}{2} + 2\cdot\sin\frac{A}{2}\cdot\cos\frac{A}{2}}{2\:\cos^{2}\frac{A}{2} + 2 \cdot\sin\frac{A}{2}\cdot \cos\frac{A}{2}} \\ &= \frac{2\:\sin\frac{A}{2}}{2\:\cos\frac{A}{2}} \cdot \Biggl(\frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}}\Biggr) \\ &= \tan\frac{A}{2} \end{align*}

$\textbf{Method 4.}$ My younger brother mentioned about this method. Multiplying numerator and deominator by $(1+\sin{A}-\cos{A})$ we get \begin{align*} \frac{1+\sin{A}-\cos{A}}{1+\sin{A}+\cos{A}} \times \frac{1+\sin{A}-\cos{A}}{1+\sin{A}-\cos{A}} &= \frac{(1+\sin{A}-\cos{A})^{2}}{(1+\sin{A})^{2}-\cos^{2}{A}} \\ &= \frac{2 + 2\: \sin{A} -2\cdot\sin{A}\cdot \cos{A} -2\cos{A}}{2\: \sin{A} + 2\sin^{2}{A}} \\ &= \frac{(2-2\cos{A}) \cdot (1+\sin{A})}{2\:\sin{A} \cdot (1+\sin{A})} \\ &=\frac{1-\cos{A}}{\sin{A}} = \frac{2\: \sin^{2}\frac{A}{2}}{2\cdot \sin\frac{A}{2} \cdot \cos\frac{A}{2}} \\ &=\tan\frac{A}{2} \end{align*}

Oh once again when I looked at the question I realize that you attempted this method. Hopefully now it's clear.

An alternative solution:

Substituting

$$\tan\left(\frac{A}{2}\right)=\frac{\sin(A)}{1+\cos(A)}$$

and multiplying through to clear the fractions quickly reduces the statement to one which is trivially true. Then working backwards through the simplifications you have made gives the proof.

Edit (providing requested clarification):

When you make the above substitution, you have

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

So

$$ (1+\sin A - \cos A)(1+\cos A) = \sin A (1+\sin A + \cos A)$$

And cancelling terms gives

$$1-\cos^2 A = \sin^2 A$$

So, using that as the starting point for your proof, "uncancelling" terms, and dividing gives the required identity.