How do you express ${sin}^{2} θ - {sec}^{2} θ + {tan}^{2} θ$ $sin^2 theta - sec^2 theta + tan^2 theta$ in terms of $cos θ$ $cos theta$ ?

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Answer 1 · 2016-06-23T05:26:46

$- {cos}^{2} x$ $-cos^2x$

${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x=1-cos^2x$

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

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

so, by substituting, you have:

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

$1-cos^2x+(cancel(-1)cancel(+1)-cos^2x)/cos^2x$

$1-cos^2x-1$

$-cos^2x$

How do you express sin^2 theta - sec^2 theta + tan^2 theta sin2θ−sec2θ+tan2θsin^2 theta - sec^2 theta + tan^2 theta in terms of cos theta cosθcos theta ?