Chapter 7.1, Problem 74E

To determine

To verify the identity:

  $\frac{\cos \theta }{1-\sin \theta }=\sec \theta +\tan \theta$

Explanation of Solution

Given:

An identity is given as

  $\frac{\cos \theta }{1-\sin \theta }=\sec \theta +\tan \theta$

Concept Used:

The trigonometric identities:

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

Algebraic identity

  $a^2-b^2=\left(a-b\right)\left(a+b\right)$

Calculation:

In order to verify the identity:

  $\frac{\cos \theta }{1-\sin \theta }=\sec \theta +\tan \theta$

Take left hand of above identity and simplify it using the trigonometric identities.

As,

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

Thus, identity is verified, i.e.

  $\frac{\cos \theta }{1-\sin \theta }=\sec \theta +\tan \theta$