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tan^2x/1+tan^2x = sin^2x
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prove rs=ls
\[\huge \frac{\tan^2{x}}{1+\tan^2{x}}=\frac{\frac{sin^2{x}}{cos^2{x}}}{1+\frac{sin^2{x}}{cos^2{x}}}=\]\[\huge \frac{\sin^2{x}}{\cos^2{x}+\sin^2{x}}=\frac{\sin^2{x}}{1}\]
any way, take 1+tan^2x as sec^2 x so, left side can be writen : tan^2x/1+tan^2x = tan^2x/sec^2 x we knowed that tanx=sinx/cosx and secx = 1/cosx => tan^2x/sec^2 x = (sin^2 x/cos^2x)/(1/cos^2 x) => tan^2x/sec^2 x = (sin^2 x/cos^2x) * cos^2 x/1 = sin^2 x
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