Question #056a7

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Answer 1 · 2017-11-05T18:04:33

${sec}^{2} (x)$ $sec^2(x)$

Rewrite the expression: $1 + tan x = 1 + \frac{sin x}{cos x}$ $1+tanx=1+sinx/cosx$

= $\frac{d}{d x} (1 + \frac{sin x}{cos x})$ $d/dx (1+sinx/cosx)$

differentiate the sum term by term

= $\frac{d}{d x} (1) + \frac{d}{d x} (\frac{sin x}{cos x})$ $d/dx(1)+d/dx(sinx/cosx)$

the derivative of 1 is zero

$\frac{d}{d x} (\frac{sin x}{cos x})$ $d/dx(sinx/cosx)$

Use the quitient rule

$\frac{\frac{d}{d x} ((sin x)) cos x - (\frac{d}{d x} (cos x) sin x)}{{cos}^{x}}$ $(d/dx((sinx))cosx-(d/dx(cosx)sinx))/(cos^(x))$

simplify

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

use the Pythagorean identity ${cos}^{2} (x) + {sin}^{2} (x) = 1$ $cos^2(x)+sin^2(x)=1$

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

$= {sec}^{2} (x)$ $=sec^2(x)$