Question #84c50

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Question #84c50

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Answer 1 · 2017-04-12T16:25:20

See below

Explanation:

$1 + \frac{{sin}^{2} θ}{{cos}^{2} θ}$ $1+(color(red)(sin^2theta))/(cos^2theta)$

Use the following identity to solve for ${sin}^{2} θ$ $sin^2theta$

${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$

${sin}^{2} θ = 1 - {cos}^{2} θ$ $sin^2theta=color(red)(1-cos^2theta)$

Substitute this instead of ${sin}^{2} θ$ $sin^2theta$

$1 + \frac{{sin}^{2} θ}{{cos}^{2} θ} = 1 + \frac{1 - {cos}^{2} θ}{{cos}^{2} θ}$ $1+(sin^2theta)/(cos^2theta)=1+(color(red)(1-cos^2theta))/cos^2theta$

Add $1$ $1$ and $\frac{1 - {cos}^{2} θ}{{cos}^{2} θ}$ $(1-cos^2theta)/cos^2theta$ together by unifying the denominators ( $1$ $1$ is

the same as $\frac{{cos}^{2} θ}{{cos}^{2} θ}$ $cos^2theta/cos^2theta$ ), so

$1 + \frac{1 - {cos}^{2} θ}{{cos}^{2} θ} = \frac{{cos}^{2} θ}{{cos}^{2} θ} + \frac{1 - {cos}^{2} θ}{{cos}^{2} θ}$ $color(red)1+(1-cos^2theta)/cos^2theta=color(red)(cos^2theta/cos^2theta)+(1-cos^2theta)/cos^2theta$

$=(cancel(cos^2theta)+1cancel(-cos^2theta))/cos^2theta$

$=1/cos^2theta$

$=sec^2theta$

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Question #84c50

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