Question

Prove that: ${\sec }^2\theta =1+{\tan }^2\theta$.

Answer 1

Let $\Delta ABP$ is a right angle triangle, where $\angle B={90}^o$.

Let $\angle PAB=\theta$.

so, $\tan \theta =\frac{PB}{AB}.....\left(1\right)$

and, $\sec \theta =\frac{PA}{AB}......\left(2\right)$

By applying Pythagoras theorem, in $\Delta ABP$,
$P{A}^2=P{B}^2+A{B}^2$
Dividing both sides by $A{B}^2$
$\Rightarrow \frac{P{A}^2}{A{B}^2}=\frac{P{B}^2}{A{B}^2}+\frac{A{B}^2}{A{B}^2}$

$\Rightarrow {\left(\frac{PA}{AB}\right)}^2={\left(\frac{PB}{AB}\right)}^2+{\left(\frac{AB}{AB}\right)}^2$ [from $\left(1\right)$ and $\left(2\right)$]

$\Rightarrow (\sec \theta {)}^2=(\tan \theta {)}^2+1$
$\Rightarrow {\sec }^2\theta ={\tan }^2\theta +1$.

$Hence,proved$.