Question

For any angle of measure $\theta$

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

Answer 1

$\begin{array}{c}\sin \theta =\frac{perpendicular}{hypo}=\frac{P}{H}\\ \cos \theta =\frac{base}{hypo}=\frac{B}{H}\\ \therefore paythaguras\\ H^2=P^2+B^2\\ \frac{H^2=P^2+B^2}{B^2}\\ \frac{H^2}{B^2}=\frac{p^2}{B^2}+\frac{B^2}{B^2}\\ 1+{\tan }^2\theta ={\sec }^2\theta \end{array}$