Let us divide both parts of the equality sin2x+cos2x=1\sin^2x+\cos^2x=1sin2x+cos2x=1 by cos2x\cos^2 xcos2x. Then we have sin2x+cos2xcos2x=1cos2x\frac{\sin^2x+\cos^2x}{\cos^2 x}=\frac{1}{\cos^2{x}}cos2xsin2x+cos2x=cos2x1 which is equivalent to sin2xcos2x+cos2xcos2x=(1cosx)2\frac{\sin^2x}{\cos^2 x}+\frac{\cos^2x}{\cos^2 x}=(\frac{1}{\cos{x}})^2cos2xsin2x+cos2xcos2x=(cosx1)2 and to tan2x+1=sec2x.\tan^2x+1=\sec^2x.tan2x+1=sec2x.
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