Chapter 7, Problem 7RE

To determine

To prove: The identity $\frac{{\cos }^{2}x}{1-\sin x}=\frac{\cos x}{\sec x-\tan x}$ .

Explanation of Solution

Given information:

The given identity is $\frac{{\cos }^{2}x}{1-\sin x}=\frac{\cos x}{\sec x-\tan x}$ .

Formula used:

Identity: $\tan x=\frac{\sin x}{\cos x}$

Reciprocal identity: $\sec x=\frac{1}{\cos x}$

Proof:

To prove the identity, take left side of the identity and simplify it.

  $\begin{array}{c}\frac{{\cos }^{2}x}{1-\sin x}=\frac{\left(\frac{{\cos }^{2}x}{\cos x}\right)}{\left(\frac{1-\sin x}{\cos x}\right)}\\ =\frac{\cos x}{\left(\frac{1}{\cos x}-\frac{\sin x}{\cos x}\right)}\\ =\frac{\cos x}{\sec x-\tan x}\left\{\begin{array}{l}\sec x=\frac{1}{\cos x}\\ \tan x=\frac{\sin x}{\cos x}\end{array}\right\}\end{array}$

From the above, it is clear that the left hand side of the identity is equal to the right side of the identity.

Hence, proved.