1. Problem: Verify and prove the identity \( \sec x = \frac{\sin 2x}{\sin x} - \frac{\cos 2x}{\cos x} \) using \( x = \frac{\pi}{4} \). 2. Recall the double-angle formulas: \[ \sin 2x = 2 \sin x \cos x, \quad \cos 2x = \cos^2 x - \sin^2 x \] 3. Substitute the double-angle formulas into the right-hand side (RHS): \[ \frac{\sin 2x}{\sin x} - \frac{\cos 2x}{\cos x} = \frac{2 \sin x \cos x}{\sin x} - \frac{\cos^2 x - \sin^2 x}{\cos x} \] 4. Simplify each term: \[ = 2 \cos x - \frac{\cos^2 x}{\cos x} + \frac{\sin^2 x}{\cos x} = 2 \cos x - \cos x + \frac{\sin^2 x}{\cos x} = \cos x + \frac{\sin^2 x}{\cos x} \] 5. Combine into a single fraction: \[ = \frac{\cos^2 x + \sin^2 x}{\cos x} = \frac{1}{\cos x} = \sec x \] 6. Since the RHS simplifies to the left-hand side (LHS), the identity is verified. --- 1. Problem: Write a description of each step in the proof of the identity: \[ \frac{\sec x}{\csc x} + \frac{\sin x}{\cos x} = 2 \tan x \] 2. Step-by-step explanation: - Line 1 to 2: Rewrite division as fraction division: \[ \frac{\sec x}{\csc x} + \frac{\sin x}{\cos x} = \sec x \div \csc x + \frac{\sin x}{\cos x} \] - Line 2 to 3: Express sec and csc in terms of sine and cosine: \[ \sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x} \] So, \[ \frac{1}{\cos x} \div \frac{1}{\sin x} + \frac{\sin x}{\cos x} \] - Line 3 to 4: Division of fractions becomes multiplication by reciprocal: \[ \frac{1}{\cos x} \times \frac{\sin x}{1} + \frac{\sin x}{\cos x} = \frac{\sin x}{\cos x} + \frac{\sin x}{\cos x} \] - Line 4 to 5: Combine like terms: \[ 2 \times \frac{\sin x}{\cos x} \] - Right side (RS) is rewritten as: \[ 2 \tan x = 2 \times \frac{\sin x}{\cos x} \] - Since LHS equals RHS, the identity is proven. Final answers: Identity verified: \( \sec x = \frac{\sin 2x}{\sin x} - \frac{\cos 2x}{\cos x} \) Proof description for \( \frac{\sec x}{\csc x} + \frac{\sin x}{\cos x} = 2 \tan x \) given above.