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Mathematics
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(sinx/(1-cosx)) + (sinx/1+cosx)) = 2 cscx
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\[\frac{sinx }{1-cosx}+\frac{sinx }{1+cosx}=\\sinx (\frac{1 }{1-cosx}+\frac{1 }{1+cosx})\\sinx (\frac{1+cosx +1-cosx }{1-\cos^2x})=\\ \frac {2\sin x}{1-\cos^2x}\] can you go on ?
1-cos^2x=sin^2x 2sinx/sin^2x =2/sinx 2/sinx= 2csc x I think I get it thanks.
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