Chapter 14.2, Problem 59HP

To determine

To Justify : The three basic Pythagorean Identities.

Explanation of Solution

Given information :

We know that, Pythagorean Identities.

Formula used :

  ${(hypotenuse)}^2={(perpendicular)}^2+{(base)}^2$

Proof :

  ${\text{In triangle ABC, b}}^{\text{2}}=c^2+a^2\mathrm{....}(1)\text{ (Using Pythagoras theorem)}$

  ${\text{Dividing by b}}^{\text{2}}$

  $\frac{b^2}{b^2}=\frac{c^2}{b^2}+\frac{a^2}{b^2}\Rightarrow 1=\frac{c^2}{b^2}+\frac{a^2}{b^2}$

  $\text{Now, sin }\theta =\frac{AB}{AC}=\frac{c}{b},\cos \theta =\frac{BC}{AC}=\frac{a}{b}$

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

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

  $Dividing(1)by{c}^2$

  $\frac{b^2}{c^2}=\frac{c^2}{c^2}+\frac{a^2}{c^2}$

  $\frac{b^2}{c^2}=1+\frac{a^2}{c^2}$

  ${\mathrm{cosec}}^2\theta =\frac{AC}{AB}=\frac{b}{c},\cot \theta =\frac{BC}{AB}=\frac{a}{c}$

  ${\mathrm{cosec}}^2\theta =1+{\cot }^2\theta$

  ${\mathrm{cosec}}^2\theta -{\cot }^2\theta =1\mathrm{.....}(3)$

  ${\text{Dividing (1) by a}}^{\text{2}}$

  $\frac{b^2}{a^2}=\frac{c^2}{a^2}+\frac{a^2}{a^2}$

  $\tan \theta =\frac{AB}{BC}=\frac{c}{a},\sec \theta =\frac{AC}{BC}=\frac{b}{a}$

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

  ${\sec }^2\theta -{\tan }^2\theta =1\mathrm{.....}(4)$

  $\text{Fromm (2), (3) and (4) Pythgorean identities are verified}\text{.}$

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

  ${\mathrm{cosec}}^2\theta -{\cot }^2\theta =1$

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