How do you prove ${sec}^{2} θ - {cos}^{2} θ {sec}^{2} θ$ $sec^2 theta - cos^2 theta sec^2 theta$ ?

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How do you prove ${sec}^{2} θ - {cos}^{2} θ {sec}^{2} θ$ $sec^2 theta - cos^2 theta sec^2 theta$ ?

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Answer 1 · 2016-06-29T13:50:51

See explanation to see if I answered your question. I proved that ${sec}^{2} θ - {cos}^{2} θ {sec}^{2} θ$ $sec^2 theta - cos^2 theta sec^2 theta$ equals ${tan}^{2} θ$ $tan^2 theta$ . Please let me know if I answered your question.

Explanation:

I'm not sure what to prove here, but I think I have an idea.

${sec}^{2} θ - {cos}^{2} θ {sec}^{2} θ$ $sec^2 theta - cos^2 theta sec^2 theta$

$sec θ$ $sec theta$ always equals $\frac{1}{cos θ}$ $1/cos theta$ (Reciprocal Identity)

${sec}^{2} θ - (\frac{{cos}^{2} θ}{1} \cdot \frac{1}{{cos}^{2} θ})$ $sec^2 theta - (cos^2 theta/1 * 1/cos^2 theta)$

$sec^2 theta - (cancel(cos^2 theta)/1 * 1/cancel(cos^2 theta))$

$sec^2 theta - (1)$

$1 + tan^2 theta$ always equals $sec^2 theta$ , or $1 + tan^2 theta = sec^2 theta$ (Pythagorean Identity)

$sec^2 theta -1$

$tan^2 theta$

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How do you prove ${sec}^{2} θ - {cos}^{2} θ {sec}^{2} θ$ $sec^2 theta - cos^2 theta sec^2 theta$ ?

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Explanation:

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How do you prove ${sec}^{2} θ - {cos}^{2} θ {sec}^{2} θ$ $sec^2 theta - cos^2 theta sec^2 theta$ ?