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:
$\displaystyle \frac{1 + \sin{x}}{1 - \sin{x}} = \left(\frac{1 + \tan{\frac{x}{2}}}{1 + \tan{\frac{x}{2}}}\right)^{2}$
is not necessarily true, since
$\displaystyle \frac{1 + \sin{x}}{1 - \sin{x}} = 1$
is not always true.
Hello cute_poisonI think you must have made a typo in your original posting, but perhaps this is what you meant:
Prove the identity $\displaystyle \frac{1+\sin x}{1-\sin x} =\left(\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right)^2$
If $\displaystyle t = \tan \left(\frac{x}{2}\right)$ then $\displaystyle \sin x = \frac{2t}{1+t^2}$. (For instance, see Half Angle Formula)
$\displaystyle \Rightarrow \frac{1+\sin x}{1-\sin x}=\frac{1+\frac{2t}{1+t^2}}{1-\frac{2t}{1+t^2}}$
$\displaystyle =\frac{(1+t^2)(1+\frac{2t}{1+t^2})}{(1+t^2)(1-\frac{2t}{1+t^2})}$
$\displaystyle =\frac{(1+t^2)+2t}{(1+t^2)-2t}$
$\displaystyle = \frac{1+2t+t^2}{1-2t+t^2}$
$\displaystyle = \frac{(1+t)^2}{(1-t)^2}$
$\displaystyle = \left(\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right)^2$
Does that help?
Grandad