Question #35a5b

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Answer 1 · 2017-04-17T18:30:38

See below

$cos x + cos x {tan}^{2} x$ $cosx+cosxtan^2x$

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

$= cos x + \frac{{sin}^{2} x}{cos x}$ $=cosx+sin^2x/cosx$

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

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

$= \frac{1}{cos x}$ $=1/cosx$

$= sec x$ $=secx$

Answer 2 · 2017-04-17T19:52:09

See below

$cos x + cos x {tan}^{2} x$ $cosx+cosxtan^2x$
$= cos x (1 + {tan}^{2} x)$ $=cosx(1+tan^2x)$
$= cos x {sec}^{2} x$ $=cosxsec^2x$
$= sec x$ $=secx$