Question

Show that the functions f and g are identically equal.

$f\left(\theta \right)=\frac{1-\sin \theta }{\cos \theta }-\frac{\cos \theta }{1+\sin \theta },g\left(\theta \right)=0$

Short answer

Yes, the functions are identically equal.

Step-by-step solution

Step 1. Given Information

The given functions are$f\left(\theta \right)=\frac{1-\sin \theta }{\cos \theta }-\frac{\cos \theta }{1+\sin \theta },g\left(\theta \right)=0$

Step 2. Calculation

Simplify any one function and determine whether they are equal or not to the other function.

$$\begin{gathered} f\left(\theta \right)=\frac{1-\sin \theta }{\cos \theta }-\frac{\cos \theta }{1+\sin \theta } \\ =\frac{(1-\sin \theta )(1+\sin \theta )-{\cos }^2\theta }{\cos \theta (1+\sin \theta )} \\ =\frac{1-{\sin }^2\theta -{\cos }^2\theta }{\cos \theta (1+\sin \theta )} \\ =\frac{1-({\sin }^2\theta +{\cos }^2\theta )}{\cos \theta (1+\sin \theta )} \\ =\frac{1-1}{\cos \theta (1+\sin \theta )} \\ =0 \\ =g\left(\theta \right) \end{gathered}$$