Chapter 12.7, Problem 6OE

To determine

Using law of sines, to find an expression equivalent to the given one.

Answer to Problem 6OE

  $\frac{\cos A}{\sin A}=\frac{\sqrt{c^2-a^2}}{a}$

Explanation of Solution

Given information :

  $\frac{\cos A}{\sin A},\angle C={90}^{\circ }$

According to law of sines,

In any triangle ABC,

  

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Hence,

  $\begin{array}{l}\frac{a}{\sin A}=\frac{c}{\sin {90}^{\circ }}\\ \sin A=\frac{a}{c}\end{array}$

Now we shall use the identity ${\sin }^{2}A+{\cos }^{2}A=1$

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

Hence,

  $\frac{\cos A}{\sin A}=\frac{\left(\frac{\sqrt{c^2-a^2}}{c}\right)}{\left(\frac{a}{c}\right)}=\frac{\sqrt{c^2-a^2}}{a}$