Question #3cd3a

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Answer 1 · 2017-03-09T16:02:42

${sec}^{2} x - \frac{{sec}^{2} x}{{csc}^{2} x}$ $sec^2x-sec^2x/csc^2x$

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

$= \frac{1}{{cos}^{2} x} - \frac{{sin}^{2} x}{{cos}^{2} x}$ $=1/cos^2x-sin^2x/cos^2x$

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

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

$= \frac{{cos}^{2} x}{{cos}^{2} x} = 1$ $=cos^2x/cos^2x=1$