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Sin(x+a)/sin(x+b)=square root of (sin2a/Sin2b) then find tanx

Sin(x+a)/sin(x+b)=square root of (sin2a/Sin2b) then find tanx

Grade:12th pass

1 Answers

Aditya Gupta
2081 Points
5 years ago
Sin2a/Sin2b= (1-cos2(x+a))/(1-cos2(x+b)) (using 1-2sin^2y=cos2y)square both the sides, you get 
cross multiply, and rearrange, you get
sin2a-sin2b=sin2acos2(x+b)-sin2bcos2(x+a)
we now multiply both sides by 2 and apply the standard formulae, to obtain
4sin(a-b)cos(a+b)=sin2(a-x-b)+sin2(x+a-b) [note that here we have used the formulae of sinx-siny, and 2sinxcosy]
this finally yields cos2x=cos(a+b)/cos(a-b)
now using tan^2x=(1-cos2x)/(1+cos2x), we get tan^2x=[cos(a-b)-cos(a+b)]/[cos(a-b)+cos(a+b)]
or tan^2x=tana*tanb
or tanx= √tana*tanb

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