Question #f37c0

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Question #f37c0

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Answer 1 · 2017-02-08T04:37:58

${sec}^{2} θ$ $sec^2theta$

Explanation:

Given expression is
$sec θ \times \frac{tan θ}{sin θ}$ $secthetaxxtantheta/sintheta$

rewriting in terms of $sin and cos$ $sin and cos$ we get
$\frac{1}{cos θ} \times \frac{sin θ}{cos θ} \times \frac{1}{sin θ}$ $1/costhetaxx(sintheta)/costhetaxx1/sintheta$
$\Rightarrow \frac{1}{cos θ} \times \frac{1}{cos θ}$ $=> 1/costhetaxx1/costheta$
$\Rightarrow \frac{1}{{cos}^{2} θ}$ $=> 1/cos^2theta$
$\Rightarrow {sec}^{2} θ$ $=> sec^2theta$

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