1. ## Prove that 3

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

2. 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,
$\displaystyle \frac{(1+n)}{(1-n)}=\frac{\tan{(\theta +\alpha)}}{\tan{\alpha}}$

3. 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

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# if sin theta= 3sin(theta 2alpha), then the value of tan(theta alpha) 2tan alpha is:

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