How do you show that $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2 theta = sec ^2 theta$ ?

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How do you show that $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2 theta = sec ^2 theta$ ?

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Answer 1 · 2018-03-03T05:14:38

Here is the answer:-

Explanation:

$1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2 theta$
$tan θ = \frac{sin θ}{cos θ}$ $tan theta=sin theta/cos theta$ and
$sec θ = \frac{1}{cos θ}$ $sec theta=1/cos theta$

$1 + \frac{{sin}^{2} θ}{{cos}^{2} θ}$ $1+sin^2theta/cos^2theta$ = $\frac{1}{{cos}^{2} θ}$ $1/cos^2theta$

${cos}^{2} θ + {sin}^{2} θ$ $cos^2theta+sin^2theta$ $/ {sin}^{2} θ$ $/sin^2theta$ = $\frac{1}{{cos}^{2} θ}$ $1/cos^2theta$

further on solving you would get both sides equal to sec^2 theta so
it is proved.

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How do you show that $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2 theta = sec ^2 theta$ ?

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How do you show that $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2 theta = sec ^2 theta$ ?