Chapter 5.1, Problem 112E

To determine

To show: The following Pythagorean identities using the triangle,

  $\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1,\\ 1+{\tan }^2\theta ={\sec }^2\theta ,\\ \text{and}\\ 1+{\cot }^2\theta ={\csc }^2\theta \end{array}$

Explanation of Solution

Given Information:

The sides of a right-angle triangle are

  $\begin{array}{l}\text{Perpendicular}=a\\ \text{Base}=b\\ \text{Hypotantenous}=\sqrt{a^2+b^2}\end{array}$

Using Pythagoras theorem in the given right-angle triangle,

  $a^2+b^2={\left(\sqrt{a^2+b^2}\right)}^2\mathrm{...}\left(1\right)$

Dividing both sides by ${\left(\sqrt{a^2+b^2}\right)}^2$ ,

  $\begin{array}{l}\frac{a^2}{{\left(\sqrt{a^2+b^2}\right)}^2}+\frac{b^2}{{\left(\sqrt{a^2+b^2}\right)}^2}=\frac{{\left(\sqrt{a^2+b^2}\right)}^2}{{\left(\sqrt{a^2+b^2}\right)}^2}\\ {\left(\frac{a}{\sqrt{a^2+b^2}}\right)}^2+{\left(\frac{b}{\sqrt{a^2+b^2}}\right)}^2=1\mathrm{...}\left(2\right)\end{array}$

Use the trigonometric formulae such that,

  $\begin{array}{l}\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotantenous}}\\ \cos \theta =\frac{\text{Base}}{\text{Hypotantenous}}\\ \tan \theta =\frac{\text{Perpendicular}}{\text{Base}}\end{array}$

Then,

  $\begin{array}{l}\sin \theta =\frac{a}{\sqrt{a^2+b^2}}\\ \cos \theta =\frac{b}{\sqrt{a^2+b^2}}\\ \tan \theta =\frac{a}{b}\end{array}$

From (2),

  $\begin{array}{l}{\left(\sin \theta \right)}^2+{\left(\cos \theta \right)}^2=1\\ {\sin }^2\theta +{\cos }^2\theta =1\end{array}$

Dividing both sides by $b^2$ in expression (1),

  $\begin{array}{l}\frac{a^2}{b^2}+\frac{b^2}{b^2}=\frac{{\left(\sqrt{a^2+b^2}\right)}^2}{b^2}\\ {\left(\frac{a}{b}\right)}^2+1={\left(\frac{\sqrt{a^2+b^2}}{b}\right)}^2\\ {\left(\frac{a}{b}\right)}^2+1={\left(\frac{1}{\frac{b}{\sqrt{a^2+b^2}}}\right)}^2\end{array}$

Therefore,

  $\begin{array}{l}{\left(\tan \theta \right)}^2+1={\left(\frac{1}{\cos \theta }\right)}^2\\ {\tan }^2\theta +1={\sec }^2\theta \end{array}$

Again, dividing both sides by $a^2$ in expression (1),

  $\begin{array}{l}\frac{a^2}{a^2}+\frac{b^2}{a^2}=\frac{{\left(\sqrt{a^2+b^2}\right)}^2}{a^2}\\ 1+{\left(\frac{b}{a}\right)}^2={\left(\frac{\sqrt{a^2+b^2}}{a}\right)}^2\\ 1+{\left(\frac{1}{\frac{a}{b}}\right)}^2={\left(\frac{1}{\frac{a}{\sqrt{a^2+b^2}}}\right)}^2\end{array}$

Therefore,

  $\begin{array}{l}1+{\left(\frac{1}{\tan \theta }\right)}^2={\left(\frac{1}{\sin \theta }\right)}^2\\ 1+{\cot }^2\theta ={\csc }^2\theta \end{array}$

Now find in the above,

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

If divide both sides by ${\sin }^2\theta$ , then

  $\begin{array}{l}\frac{{\sin }^2\theta }{{\sin }^2\theta }+\frac{{\cos }^2\theta }{{\sin }^2\theta }=\frac{1}{{\sin }^2\theta }\\ 1+{\cot }^2\theta ={\csc }^2\theta \end{array}$

If divide both sides by ${\cos }^2\theta$ , then

  $\begin{array}{l}\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{{\cos }^2\theta }{{\cos }^2\theta }=\frac{1}{{\cos }^2\theta }\\ {\tan }^2\theta +1={\sec }^2\theta \end{array}$

Conclusion:

  $\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\\ 1+{\cot }^2\theta ={\csc }^2\theta \\ {\tan }^2\theta +1={\sec }^2\theta \end{array}$