Answer
See the full explanation below.
Work Step by Step
Let us consider the left side of the given expression:
$\frac{\sec x-\csc x}{\sec x+\csc x}$
Apply the reciprocal identity of trigonometry $\sec x=\frac{1}{\cos x}$ , and $\csc x=\frac{1}{sinx}$ , then the above expression can be further simplified as:
$\frac{\sec x-\csc x}{\sec x+\csc x}=\frac{\frac{1}{\cos x}-\frac{1}{\sin x}}{\frac{1}{\cos x}+\frac{1}{\sin x}}$
We multiply the numerator and denominator by $sinx$.
$\begin{align}
& \frac{\frac{1}{\cos x}-\frac{1}{\sin x}}{\frac{1}{\cos x}+\frac{1}{\sin x}}=\frac{\frac{1}{\cos x}-\frac{1}{\sin x}}{\frac{1}{\cos x}+\frac{1}{\sin x}}.\frac{\sin x}{\sin x} \\
& =\frac{\frac{\sin x}{\cos x}-\frac{\sin x}{\sin x}}{\frac{\sin x}{\cos x}+\frac{\sin x}{\sin x}} \\
& =\frac{\frac{\sin x}{\cos x}-1}{\frac{\sin x}{\cos x}+1}
\end{align}$
By using the quotient identity of trigonometry $\tan x=\frac{\sin x}{\cos x}$, the above expression can be further simplified as:
$\frac{\frac{\sin x}{\cos x}-1}{\frac{\sin x}{\cos x}+1}=\frac{\tan x-1}{\tan x+1}$
Hence, the left side of the expression is equal to the right side, which is
$\frac{\sec x-\csc x}{\sec x+\csc x}=\frac{\tan x-1}{\tan x+1}$.
