If sin`theta=n"sin"(theta+2alpha)` , prove that `tan(theta+alpha)=(1+n)/(1-n)tanalpha` .

Explanation

Given

`sintheta=nsin(theta+2alpha)`

`=>(sintheta)/(sin(theta+2alpha))=n
`
apply componendo dividendo rule,

`=>(sintheta+sin(theta+2alpha))/(sintheta−sin(theta+2alpha))=(n+1)/(n-1)`

applying `sinC+sinD=2sin((C+D)/2)cos((C-D)/2)` and `sinC−sinD=2sin((C-D)/2)cos((C+D)/2)` formula

`=>(2sin(theta+alpha)cosalpha)/(2sinalphacos(theta+alpha))=(n+1)/(n-1)`

`=>(tan(theta+alpha))/(tanalpha)=(n+1)/(n-1)`

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