if sintheta = nsin(theta + 2 alpha)
prove that tan(theta + alpha) = (1+n)/(1-n) *tan theta
my incomplete soln -
sin theta/ sin(theta+ 2 alpha) = n
I guess you've made some mistake here...
I think what we have to prove is that $\displaystyle \tan (\theta +\alpha) = \frac{(1+n)}{(1-n)}\tan \alpha $
if so, take $\displaystyle \frac{(1+n)}{(1-n)}$
for this you'll need the following equation,
$\displaystyle \sin C +\sin D = 2\sin{\frac{(C+D)}{2}}.\cos{\frac{(C-D)}{2}}$
eventually you'll get,
$\displaystyle \frac{(1+n)}{(1-n)}=\frac{\tan{(\theta +\alpha)}}{\tan{\alpha}}$
sin(theta)/sin(theta+2alpha)=n/1
sin(theta+2alpha) + sin(theta) / sin(theta+2alpha) - sin(theta) = 1+n/1-n
using sin P + sin Q = 2sin[(P+Q)/2] cos[(P-Q)/2] and sin p - sin Q = 2cos[(P+Q)/2] sin[(P-Q)/2]
we get tan(theta + alpha)/tan( alpha) = 1+n/1-n and hence required result