# Thread: trig proof

1. ## trig proof

1+sinx/1-sinx=((1+tan(x/2))/(1+tan(x/2))^2

Need to prove it but cant get my head round it!!if anyone could help asap!!thanks!!

2. Originally Posted by cute_poison
1+sinx/1-sinx=((1+tan(x/2))/(1+tan(x/2))^2

Need to prove it but cant get my head round it!!if anyone could help asap!!thanks!!

Ok, before I work on this, let me make sure I am working the problem you have because this:

$\frac{1 + \sin{x}}{1 - \sin{x}} = \left(\frac{1 + \tan{\frac{x}{2}}}{1 + \tan{\frac{x}{2}}}\right)^{2}$

is not necessarily true, since

$\frac{1 + \sin{x}}{1 - \sin{x}} = 1$

is not always true.

3. Originally Posted by sinewave85
Ok, before I work on this, let me make sure I am working the problem you have because this:

$\frac{1 + \sin{x}}{1 - \sin{x}} = \left(\frac{1 + \tan{\frac{x}{2}}}{1 + \tan{\frac{x}{2}}}\right)^{2}$

is not necessarily true, since

$\frac{1 + \sin{x}}{1 - \sin{x}} = 1$

is not always true.
Your RHS simplifies to 1 so I'd imagine the OP meant

$\frac{1 + \sin{x}}{1 - \sin{x}} = \frac{1+tan({\frac{x}{2}})}{1+tan^2({\frac{x}{2}}) }$

4. Originally Posted by e^(i*pi)
Your RHS simplifies to 1 so I'd imagine the OP meant

$\frac{1 + \sin{x}}{1 - \sin{x}} = \frac{1+tan({\frac{x}{2}})}{1+tan^2({\frac{x}{2}}) }$
Yeah, I read it wrong, thanks.

5. I hope what the op meant was:

$\frac{1 + \sin{x}}{1 - \sin{x}} = \frac{1+tan({\frac{x}{2}})}{[1+tan({\frac{x}{2}})]^{2}}
$

since that is how I think it reads.
(The square is outside the brackets for the right-hand denominator).

6. In either case, you are going to need the half-angle identities for tan, which are:

$\tan{\frac{x}{2}} = \frac{1 - \cos{x}}{\sin{x}}$

$\tan{\frac{x}{2}} = \frac{\sin{x}}{1 + \cos{x}}$

7. ## Trig Identity

Hello cute_poison
Originally Posted by cute_poison
1+sinx/1-sinx=((1+tan(x/2))/(1+tan(x/2))^2

Need to prove it but cant get my head round it!!if anyone could help asap!!thanks!!
I think you must have made a typo in your original posting, but perhaps this is what you meant:

Prove the identity $\frac{1+\sin x}{1-\sin x} =\left(\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right)^2$

If $t = \tan \left(\frac{x}{2}\right)$ then $\sin x = \frac{2t}{1+t^2}$. (For instance, see Half Angle Formula)

$\Rightarrow \frac{1+\sin x}{1-\sin x}=\frac{1+\frac{2t}{1+t^2}}{1-\frac{2t}{1+t^2}}$

$=\frac{(1+t^2)(1+\frac{2t}{1+t^2})}{(1+t^2)(1-\frac{2t}{1+t^2})}$

$=\frac{(1+t^2)+2t}{(1+t^2)-2t}$

$= \frac{1+2t+t^2}{1-2t+t^2}$

$= \frac{(1+t)^2}{(1-t)^2}$

$= \left(\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right)^2$

Does that help?