Prove the identity \[ \frac{\sec x+1}{\tan x}=\frac{\tan x}{\sec x-1} \]

To prove the identity \[ \frac{\sec x + 1}{\tan x} = \frac{\tan x}{\sec x - 1}, \] we start by rewriting the left-hand side. Recall that \(\sec x = \frac{1}{\cos x}\) and \(\tan x = \frac{\sin x}{\cos x}\). Thus, the left-hand side can be expressed as: \[ \frac{\sec x + 1}{\tan x} = \frac{\frac{1}{\cos x} + 1}{\frac{\sin x}{\cos x}} = \frac{\frac{1 + \cos x}{\cos x}}{\frac{\sin x}{\cos x}}. \] This simplifies to: \[ \frac{1 + \cos x}{\sin x}. \] Now, let’s simplify the right-hand side: \[ \frac{\tan x}{\sec x - 1} = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x} - 1} = \frac{\frac{\sin x}{\cos x}}{\frac{1 - \cos x}{\cos x}}. \] This simplifies to: \[ \frac{\sin x}{1 - \cos x}. \] Now we have the two expressions: 1. Left-hand side: \(\frac{1 + \cos x}{\sin x}\) 2. Right-hand side: \(\frac{\sin x}{1 - \cos x}\) Next, we can cross-multiply to set them equal: \[ (1 + \cos x)(1 - \cos x) = (\sin x)(\sin x). \] Expanding both sides: The left side becomes: \[ 1 - \cos^2 x = \sin^2 x, \] which is indeed a true statement since \(1 - \cos^2 x = \sin^2 x\). Thus, both sides are equal, and we have shown that: \[ \frac{\sec x + 1}{\tan x} = \frac{\tan x}{\sec x - 1}, \] proving the identity is true.