Chapter 13.2, Problem 52HP

To determine

To find : Which equation does not belong to other three.

Answer to Problem 52HP

  ${\sin }^2\theta -{\cos }^2\theta =2{\sin }^2\theta$

Explanation of Solution

Given information :

We know that, $\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1,{\sin }^2\theta -{\cos }^2\theta =2{\sin }^2\theta ,1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta \\ and{\tan }^2\theta +1={\sec }^2\theta \end{array}$

Calculation :

As $\sin \theta =\frac{\text{Perpendicular}}{\text{Hypoteneus}},\cos \theta =\frac{\text{Base}}{\text{Hypoteneus}},\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}$

In a right angle triangle.

  ${\sec }^2\theta -{\cos }^2\theta =2{\sin }^2\theta$ is not a Pythagorean identity as the other three are.

So, ${\text{sin}}^{\text{2}}\theta -{\cos }^2\theta =2{\sin }^2\theta$ does not belong to other three.