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`        if Sin(x+a)/Sin(x+b)=square root of (Sin2a/Sin2b) then tanx`
one year ago

```							ohh! how much i wish askiitians had an attachment feature just like quora :(, anyways, i mma write it anyways.so, square both the sides, you get Sin2a/Sin2b= (1-cos2(x+a))/(1-cos2(x+b)) (using 1-2sin^2y=cos2y)cross multiply, and rearrange, you getsin2a-sin2b=sin2acos2(x+b)-sin2bcos2(x+a)we now multiply both sides by 2 and apply the standard formulae, to obtain4sin(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*tanbor tanx= √tana*tanb
```
one year ago
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