Question

Show that the functions f and g are identically equal.

$f\left(\theta \right)=\tan \theta +sec\theta ,g\left(\theta \right)=\frac{\cos \theta }{1-\sin \theta }$

Short answer

Yes, the functions are identically equal.

Step-by-step solution

Step 1. Given Information

The given functions are$f\left(\theta \right)=\tan \theta +sec\theta ,g\left(\theta \right)=\frac{\cos \theta }{1-\sin \theta }$

Step 2. Calculation

Simplify both the functions and determine whether they are equal or not to each other.

$$\begin{gathered} f\left(\theta \right)=\tan \theta +sec\theta \\ =\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta } \\ =\frac{1+\sin \theta }{\cos \theta } \end{gathered}$$

And

$$\begin{gathered} g\left(\theta \right)=\left(\frac{\cos \theta }{1-\sin \theta }\right)\left(\frac{1+\sin \theta }{1+\sin \theta }\right) \\ =\frac{\cos \theta (1+\sin \theta )}{(1-\sin \theta )(1+\sin \theta )} \\ =\frac{\cos \theta (1+\sin \theta )}{1-{\sin }^2\theta } \\ =\frac{\cos \theta (1+\sin \theta )}{{\cos }^2\theta } \\ =\frac{1+\sin \theta }{\cos \theta } \end{gathered}$$

Thus, both the functions are equal.