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Verify the identity.
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\[1 + \sec ^{2}xsin ^{2}x = \sec ^{2}x\]
\[\frac{ 1 }{ \cos^2x }=\sec^2x\]
is there a shorter way of explaning it.
\[se^2 =\frac{1}{cos^2}\\therefore1+\frac{sin^2}{cos^2}=1+tan^2=sec^2\]
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using \[\sin^2x+\cos^2=1\] from second step: \[\frac{ \sin^2x }{\cos^2x }+\frac{ \cos^2x }{\cos^2x }=\frac{ 1 }{ \cos^2x }\] \[\tan^2x+1=\sec^2x\]
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