Chapter 5.2, Problem 62E

Solution

To choose: The correct option from the given options.

The correct option is (C).

Given information:

The given identity is $\tan \theta +\sec \theta =\frac{\cos \theta }{1-\sin \theta }$ .

Formula used:

The trigonometric identity is $1-{\sin }^{2}x={\cos }^{2}x$ .

The algebraic identity $\left(a-b\right)\left(a+b\right)=a^2-b^2$ .

Explanation:

Rewrite the left side of the given identity in terms of $\sin \theta$ and $\cos \theta$ , and simplify using algebraic and trigonometric identities.

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

Now, apply the identity $1-{\sin }^{2}x={\cos }^{2}x$ and simplify the obtained expression.

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

This implies that the required intermediate step to prove the identity is $\frac{\sin \theta +1}{\cos \theta }$ .

Therefore, the correct option is (C).