How do you prove that ${sec}^{2} x + {csc}^{2} x = 1$ $sec^2x+csc^2x=1$ is not an identity?

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Answer 1 · 2016-11-12T12:52:44

${sec}^{2} x + {csc}^{2} x = \frac{1}{{sin}^{2} x {cos}^{2} x}$ $sec^2x+csc^2x=1/(sin^2xcos^2x)$

Since $sec x = \frac{1}{cos x}$ $secx=1/cosx$ and $csc x = \frac{1}{sin x}$ $cscx=1/sinx$ , you will get:

${sec}^{2} x + {csc}^{2} x = \frac{1}{{cos}^{2} x} + \frac{1}{{sin}^{2} x}$ $sec^2x+csc^2x=1/cos^2x+1/sin^2x$

$= \frac{{sin}^{2} x + {cos}^{2} x}{{sin}^{2} x {cos}^{2} x}$ $=(sin^2x+cos^2x)/(sin^2xcos^2x)$

$= \frac{1}{{sin}^{2} x {cos}^{2} x}$ $=1/(sin^2xcos^2x)$

How do you prove that sec2x+csc2x=1sec^2x+csc^2x=1 is not an identity?