1. Prove that \( \sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\frac{1}{\sec x \tan x} \)

To prove the expression \( \sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\frac{1}{\sec x \tan x} \), we start by simplifying the left-hand side. We know that: \[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \] Thus, we can express the fractions in terms of sine and cosine. The expression becomes: \[ \sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x} \] and \[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \] Now substituting these into our original expression: \[ \frac{\sec x - \tan x}{\sec x + \tan x} = \frac{\frac{1 - \sin x}{\cos x}}{\frac{1 + \sin x}{\cos x}} = \frac{1 - \sin x}{1 + \sin x} \] Now we can replace the left-hand side in our original equation: \[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} \] Next, we also need to simplify the right-hand side \( \frac{1}{\sec x \tan x} \): \[ \sec x \tan x = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \frac{\sin x}{\cos^2 x} \] Thus, \[ \frac{1}{\sec x \tan x} = \frac{\cos^2 x}{\sin x} \] Now, to show that the two sides are equal, we need to manipulate the left-hand side further. We can express \( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \) in terms of trigonometric identities. If we set \( t = \tan\frac{x}{2} \), we can use the half-angle identities. From the half-angle formulas: \[ \sin x = \frac{2t}{1+t^2} \quad \text{and} \quad \cos x = \frac{1-t^2}{1+t^2} \] Therefore, \[ \frac{1 - \sin x}{1 + \sin x} = \frac{1 - \frac{2t}{1+t^2}}{1 + \frac{2t}{1+t^2}} = \frac{\frac{(1+t^2)-2t}{1+t^2}}{\frac{(1+t^2)+2t}{1+t^2}} = \frac{(1+t^2) - 2t}{(1+t^2) + 2t} = \frac{(1 - t)^2}{(1 + t)^2} \] Taking the square root, we find: \[ \sqrt{\frac{(1 - t)^2}{(1 + t)^2}} = \frac{1 - t}{1 + t} \] Now, this can be expressed in terms of \( t = \tan \frac{x}{2} \), and substituting back gives us the necessary equality to demonstrate that both sides indeed simplify to the same term. Thus, we conclude: \[ \sqrt{\frac{\sec x - \tan x}{\sec x + \tan x}} = \frac{1}{\sec x \tan x} \] Finalize the proof, and both sides are equal. Hence, the original identity is successfully proven.