Chapter 5.2, Problem 90E

(a)

To determine

To prove: The given identity.

Explanation of Solution

Given information:

The following identity:

  $\frac{\sec x}{\tan x}=\frac{\tan x}{\sec x-\cos x}$

Formula used:

The following identities:

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

Proof:

The given identity can be verified as follows:

  $\frac{\sec x}{\tan x}=\frac{\tan x}{\sec x-\cos x}\text{ [Given]}$

Working with the left side,

  $\begin{array}{l}\frac{\sec x}{\tan x}\\ =\frac{\frac{1}{\cos x}}{\frac{\sin x}{\cos x}}\text{ [Reciprocal identity and quotient identity]}\\ =\frac{1}{\cos x}\cdot \frac{\cos x}{\sin x}\text{ [Multiply by reciprocal of denominator]}\\ =\frac{1}{\sin x}\text{ [Simplify]}\\ =\csc x\text{ [Reciprocal identity]}\end{array}$

Working with the right side

  $\begin{array}{l}\frac{\tan x}{\sec x-\cos x}\\ =\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}-\cos x}\text{ [Quotient identity and reciprocal identity]}\\ =\frac{\frac{\sin x}{\cos x}}{\frac{1-{\cos }^{2}x}{\cos x}}\text{ [Subtract the fractions in the denominator]}\\ =\frac{\frac{\sin x}{\cos x}}{\frac{{\sin }^{2}x}{\cos x}}\text{ [Pythagorean identity]}\\ =\frac{\sin x}{\cos x}\cdot \frac{\cos x}{{\sin }^{2}x}\text{ [Multiply by reciprocal of denominator]}\\ =\frac{1}{\sin x}\text{ [Simplify]}\\ =\csc x\text{ [Reciprocal identity]}\end{array}$

The identity is verified because both sides are equal to $\csc x$ .

(b)

To determine

To prove: If the given identity is true for the given value of $x$ .

Explanation of Solution

Given information:

The following identity:

  $\frac{\sec x}{\tan x}=\frac{\tan x}{\sec x-\cos x}$

The following value:

  $x=\pi$

Proof:

The prove if the identity is true for the given value of $x$ , substitute the value of $x$ in the identity.

Working with the left side,

  $\begin{array}{l}\frac{\sec x}{\tan x}\\ =\frac{\sec 180}{\tan 180}\text{ [Substitute value of }x]\\ =\frac{1}{0}\text{ [Put values]}\\ =\infty \text{ [Divide]}\end{array}$

Working with the right side

  $\begin{array}{l}\frac{\tan x}{\sec x-\cos x}\\ =\frac{\tan 180}{\sec 180-\cos 180}\text{ [Substitute value of }x]\\ =\frac{0}{-1-(-1)}\text{ [Put values]}\\ =\frac{0}{-1+1}\text{ [Simplify]}\\ \text{=}\frac{0}{0}\text{ [Add] }\\ \text{= }\infty \text{ [Divide]}\end{array}$

The identity is true for $x=\pi$ because both sides are equal to $\infty$ .