Question:
If $\cos \theta=\frac{2}{3}$, find the value of $\frac{\sec \theta-1}{\sec \theta+1}$
Solution:
Given in question: $\cos \theta=\frac{2}{3}$
We have to find $\frac{\sec \theta-1}{\sec \theta+1}$
$\Rightarrow \frac{\sec \theta-1}{\sec \theta+1}=\frac{\frac{1}{\cos \theta}-1}{\frac{1}{\cos \theta}+1}$
$\Rightarrow \frac{\sec \theta-1}{\sec \theta+1}=\frac{\frac{3}{2}-1}{\frac{3}{2}+1}$
$\Rightarrow \frac{\sec \theta-1}{\sec \theta+1}=\frac{1}{5}$
Hence the value of $\frac{\sec \theta-1}{\sec \theta+1}$ is $\frac{1}{5}$