# If cos θ=23, find the value of sec θ−1sec θ+1.

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 $\frac{1}{5}$