Question

Square relations among various trigonometrical ratios
${\sin }^2\theta +{\cos }^2\theta =1$
${\sec }^2\theta -{\tan }^2=1$
$\cos e{c}^2\theta -{\cot }^2\theta =1$
These are also called Pythagorean identities.

Answer 1

$\begin{array}{c}Let\Delta ABC\\ \angle ABC={90}^{\circ }\\ {\sin }^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2=1\\ cose{c}^2\theta -co{t}^2\theta =1\\ \\ Sin\theta =\frac{p}{h},\cos \theta =\frac{b}{h}\\ byu\sin gpythagorastheoram\\ \Rightarrow P^2+b^2=h^2\\ \Rightarrow \frac{p^2}{h^2}+\frac{b^2}{h^2}=1\\ {\sin }^2\theta +{\cos }^2\theta =1\left(proved\right)\\ \Rightarrow {\sec }^2\theta -{\tan }^2=1\\ \Rightarrow \frac{h^2}{b^2}+\frac{p^2}{b^2}=1\\ \Rightarrow \frac{h^2+p^2}{b^2}=1\\ \Rightarrow \frac{b^2}{b^2}=1\left(proved\right)\\ and\\ cose{c}^2\theta -co{t}^2\theta =1\\ \Rightarrow \frac{h^2}{p^2}-\frac{b^2}{p^2}=1\\ \Rightarrow \frac{h^2-b^2}{p^2}=1\\ \Rightarrow \frac{p^2}{p^2}=1\left(proved\right)\end{array}$