sinx/(1+cosx)
simplify that so it equals:
tan(x/2)
(1-cosx)/sinx
simplify that so it equals:
tan(x/2)
so can someone post the steps please and thanks
start with the basic ratio identity for tangent ...
$\displaystyle \tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}$
you will also need the half-angle identities for sine ...
$\displaystyle \sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1-\cos{x}}{2}}$
and cosine ...
$\displaystyle \cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1+\cos{x}}{2}}$
Use the double angle identities to express sin and cos in terms of their half-angles since $\displaystyle \sin(x) = \sin \left[2\left(\frac{x}{2}\right)\right]$
Let $\displaystyle \theta = \dfrac{x}{2}$ and we arrive at:
$\displaystyle \dfrac{2\sin(\theta)\cos(\theta)}{1+2\cos^2 (\theta) -1} = \dfrac{2\sin \theta \cos \theta}{2\cos^2 \theta}$
$\displaystyle \sin (x) = \sin \left( {2\cdot\tfrac{x}{2}} \right) = 2\sin \left( {\tfrac{x}{2}} \right)\cos \left( {\tfrac{x}{2}} \right)$.
In a similar way we get $\displaystyle \cos (x) = \cos \left( {2\cdot\tfrac{x}{2}} \right) = 2\cos^2 \left( {\tfrac{x}{2}} \right)-1$.
Use those two in $\displaystyle \frac{\sin(x)}{\cos(x)+1}$ it falls right out.