Question #3a93e

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Question #3a93e

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Answer 1 · 2017-05-07T18:49:41

Use Pythagorean identity and definition of tangent

Explanation:

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

Remember, $tan (x) = \frac{sin (x)}{cos (x)}$ $tan(x)=sin(x)/cos(x)$ (this is just the definition of $tan (x)$ $tan(x)$ )

${tan}^{2} (x) + 1$ $tan^2(x)+1$

$= {(tan (x))}^{2} + 1$ $=(tan(x))^2+1$

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

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

Common denominator

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

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

Now we use the Pythagorean identity ( $1 = {cos}^{2} (x) + {sin}^{2} (x)$ $1=cos^2(x)+sin^2(x)$ , I won't prove this here, but if you want, check this out) to solve the rest

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

There it is,

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

Answer 2 · 2017-05-07T20:20:10

We know by definition
$cos θ = \frac{x}{r}$ $costheta=x/r$
$tan θ = \frac{y}{x}$ $tantheta=y/x$
enter image source here
Now,
$1 + {tan}^{2} θ = 1 + \frac{y^{2}}{x^{2}}$ $1+tan^2theta=1+y^2/x^2$
$= \frac{x^{2} + y^{2}}{x^{2}}$ $=(x^2+y^2)/x^2$
$= \frac{r^{2}}{x^{x}}$ $=r^2/x^x$
$= \frac{1}{{cos}^{2} θ}$ $=1/cos^2theta$

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Question #3a93e

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