1. ## showing identity help..

Hey guys, need help showing this identity:

$\displaystyle \frac{1+sinx-cosx}{-1+sinx+cosx} = \frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}$

any help much appreciated!

2. Originally Posted by jvignacio
Hey guys, need help showing this identity:

$\displaystyle \frac{1+sinx-cosx}{-1+sinx+cosx} = \frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}$

any help much appreciated!

The $\displaystyle \tan x$ half-angle identities are: $\displaystyle \tan \left(\frac{x}{2}\right) = \csc x - \cot x = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{\sin x}{1 + \cos x} = \frac{1-\cos x}{\sin x}$

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

Consider one side and equate to the other thus: $\displaystyle \mathrm{RHS} = \frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)} = \frac{1+\left(\frac{1-\cos x}{\sin x}\right) }{1-\left(\frac{1-\cos x}{\sin x}\right)}= \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \mathrm{LHS}$

3. Originally Posted by Air
The $\displaystyle \tan x$ half-angle identities are: $\displaystyle \tan \left(\frac{x}{2}\right) = \csc x - \cot x = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{\sin x}{1 + \cos x} = \frac{1-\cos x}{\sin x}$

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

Consider one side and equate to the other thus: $\displaystyle \mathrm{RHS} = \frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)} = \frac{1+\left(\frac{1-\cos x}{\sin x}\right) }{1-\left(\frac{1-\cos x}{\sin x}\right)}= \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \mathrm{LHS}$
thanks for that!

for this one: $\displaystyle \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \mathrm{LHS}$

did you just cancel out the sinx in the top fraction with the buttom fraction and re arrange to make the 1's first?

4. Originally Posted by jvignacio
thanks for that!

for this one: $\displaystyle \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \mathrm{LHS}$

did you just cancel out the sinx in the top fraction with the buttom fraction and re arrange to make the 1's first?
These are the steps in between (in the end, I just rearranged to get in in the same format as the question asked):
$\displaystyle \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}} = \frac{\sin x + 1-\cos x}{\sin x} \div \frac{\sin x - 1+\cos x}{\sin x} =$ $\displaystyle \frac{\sin x + 1-\cos x}{\sin x} \times \frac{\sin x}{\sin x - 1+\cos x}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x}$

5. Originally Posted by Air
These are the steps in between (in the end, I just rearranged to get in in the same format as the question asked):
$\displaystyle \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}} = \frac{\sin x + 1-\cos x}{\sin x} \div \frac{\sin x - 1+\cos x}{\sin x} =$ $\displaystyle \frac{\sin x + 1-\cos x}{\sin x} \times \frac{\sin x}{\sin x - 1+\cos x}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x}$
yes yes now i understand. thanks for that