Question:
If $\tan \theta=\frac{a}{b}$, find the value of $\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}$.
Solution:
Given:
$\tan \theta=\frac{a}{b} \ldots \ldots$(1)
Now, we know that $\tan \theta=\frac{\sin \theta}{\cos \theta}$
Therefore equation (1) becomes as follows
$\frac{\sin \theta}{\cos \theta}=\frac{a}{b}$
Now, by applying invertendo
We get,
$\frac{\cos \theta}{\sin \theta}=\frac{b}{a}$
Now, by applying Compenendo-dividendo
We get,
$\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{b+a}{b-a}$
Therefore,
$\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{b+a}{b-a}$