Chapter SH, Problem 13.9EP

To determine

To prove:

The each expression into trigonometric form

Explanation of Solution

Given:

  $\text{s}{\text{in}}^2\theta +{\cos }^2\theta \text{=1}$

Concept used:

Trigonometric ratios: - Atrigonometric ratio are the ratios between edges of a right triangle.

Calculation:

By Pythagoras theorem

  ${\left(AB\right)}^2+{\left(BC\right)}^2={\left(AC\right)}^2\mathrm{..............}\left(1\right)$

On dividing equation (1) with ${\left(AC\right)}^2$

  $\begin{array}{l}\frac{{\left(AB\right)}^2}{{\left(AC\right)}^2}+\frac{{\left(BC\right)}^2}{{\left(AC\right)}^2}=\frac{{\left(AC\right)}^2}{{\left(AC\right)}^2}\\ \frac{{\left(AB\right)}^2}{{\left(AC\right)}^2}+\frac{{\left(BC\right)}^2}{{\left(AC\right)}^2}=1\\ \text{s}{\text{in}}^2\theta +{\cos }^2\theta \text{=1}\end{array}$