Chapter 5.1, Problem 69E

To determine

To express: The given trigonometric expression in non-fractional form.

Answer to Problem 69E

The expression $\frac{3}{\sec x-\tan x}$ is equal to $[3(\sec x+\tan x)]$

Explanation of Solution

Given information:

The given trigonometric expression:

  $\frac{3}{\sec x-\tan x}$

Formula used:

Pythagorean identity:

  ${\cos }^{2}x=1-{\sin }^{2}x=(1+\sin x)(1-\sin x)$

Calculation:

As per problem,

The given expression:

  $\frac{3}{\sec x-\tan x}$

From the Pythagorean identity: ${\cos }^{2}x=1-{\sin }^{2}x=(1+\sin x)(1-\sin x)$

Multiply both numerator and denominator by $(1+\sin x)$ after writing the expression in terms of sine and cosine.

  $\begin{array}{l}\frac{3}{\sec x-\tan x}=\frac{3}{\frac{1}{\cos x}-\frac{\sin x}{\cos x}}\text{ [reciprocal identity: }\sec x=\frac{1}{\cos x}\text{ and quotient identity:}\tan x=\frac{\sin x}{\cos x}\text{] }\\ \text{ =}\frac{3\cos x}{1-\sin x}\text{ [subtract fractions in the denominator and simplify]}\\ \text{ =}\frac{3\cos x}{1-\sin x}\cdot \frac{(1+\sin x)}{(1+\sin x)}\text{ [multiply numerator and denominanator by (1+}\sin x\text{)]}\\ \text{ =}\frac{3\cos x(1+\sin x)}{1-{\sin }^{2}x}\text{ [multiply]}\\ \text{ =}\frac{3\cos x(1+\sin x)}{{\cos }^{2}x}\text{ [pythagorean identity]}\\ \text{ =}\frac{3(1+\sin x)}{\cos x}\text{ [cancel out common factors and simplify]}\\ \text{ =3}\left(\frac{1}{\cos x}\text{+}\frac{\sin x}{\cos x}\right)\text{ [write as separate fractions]}\\ \text{ =3}\left(\sec x\text{+}\tan x\right)\text{ [reciprocal identity and quotient identity]}\end{array}$

Hence, the expression $\frac{3}{\sec x-\tan x}$ is equal to $[3(\sec x+\tan x)]$ .