Chapter 12.2, Problem 31WE

To determine

Use this first Pythagorean identity to prove that the other two identities are true for every acute angle $\theta$ .

Explanation of Solution

Given:

The three Pythagorean identities in trigonometry are as follows:

  ${\sin }^2\theta +{\cos }^2\theta =1$

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

  $1+{\cot }^2\theta ={\csc }^2\theta$

Formula Used:

  $\tan \theta =\frac{\sin \theta }{\cos \theta }$

  $\cot \theta =\frac{\cos \theta }{\sin \theta }$

  $\csc \theta =\frac{1}{\sin \theta }$

  $\sec \theta =\frac{1}{\cos \theta }$

Calculation:

First identity:

  ${\sin }^2\theta +{\cos }^2\theta =\mathrm{1..............}(1)$

Divide each side of equation (1) by ${\cos }^2\theta$

  $\begin{array}{l}\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{{\cos }^2\theta }{{\cos }^2\theta }=\frac{1}{{\cos }^2\theta }\\ \text{⇒}{\tan }^2\theta +1={\sec }^2\theta \mathrm{..............}[\tan \theta =\frac{\sin \theta }{\cos \theta },\sec \theta =\frac{1}{\cos \theta }]\end{array}$

Divide each side of equation (1) by ${\sin }^2\theta$

  $\begin{array}{l}\frac{{\sin }^2\theta }{{\sin }^2\theta }+\frac{{\cos }^2\theta }{{\sin }^2\theta }=\frac{1}{{\sin }^2\theta }\\ \text{⇒}1+{\cot }^2\theta ={\csc }^2\theta \mathrm{..............}[\cot \theta =\frac{\cos \theta }{\sin \theta },\csc \theta =\frac{1}{\sin \theta }]\end{array}$

Hence proved.