How do you prove $1 + {tan}^{2} (x) = {sec}^{2} (x)$ $1+tan^2 (x) = sec^2 (x)$ ?

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Answer 1 · 2016-05-02T21:37:32

Using the following:

for $cos (x) \neq 0$ $cos(x)!=0$ , we have

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

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

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

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

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

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

How do you prove 1+tan2(x)=sec2(x)1+tan^2 (x) = sec^2 (x)?