Question #8d661

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Question #8d661

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Answer 1 · 2017-06-22T08:23:22

$sec x$ $secx$

Explanation:

$\frac{1}{sec x} + \frac{{tan}^{2} x}{sec x}$ $1/secx+tan^2x/secx$
= $\frac{1 + {tan}^{2} x}{sec x}$ $(1+tan^2x)/secx$

Using the identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$
and hence the derivation from dividing both sides by ${cos}^{2} x$ $cos^2x$ : $1 + {tan}^{2} x = {sec}^{2} x$ $1+tan^2x=sec^2x$
$\frac{1 + {tan}^{2} x}{sec x}$ $(1+tan^2x)/secx$ becomes $\frac{{sec}^{2} x}{sec x}$ $sec^2x/secx$
which further simplifies into $sec x$ $secx$

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Question #8d661

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