Question

Simplify: \( \tan x - \cot x + \csc x \sec x \)

• A. \( 2 \tan x \)
• B. \( 2 \csc x \sec x \)
• C. \( 2 \tan x \sec x \)
• D. \( 2 \cot x \)
• E. \( 2 \cot x \csc x \)

Hint:

Convert trigonometric functions into sine and cosine before simplifying.

Answer 1

The Correct Option is A

Step 1: Express in Terms of Sine and Cosine \[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x}, \quad \csc x = \frac{1}{\sin x}, \quad \sec x = \frac{1}{\cos x} \] Step 2: Compute the Given Expression \[ \tan x - \cot x + \csc x \sec x \] \[ = \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} + \frac{1}{\sin x} \times \frac{1}{\cos x} \] \[ = \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} + \frac{1}{\sin x \cos x} \] Step 3: Factorize \[ = \frac{\sin^2 x - \cos^2 x + 1}{\sin x \cos x} \] Since \( \sin^2 x + \cos^2 x = 1 \), we get: \[ = \frac{1 - \cos^2 x + \sin^2 x}{\sin x \cos x} = \frac{\sin^2 x + \sin^2 x}{\sin x \cos x} \] \[ = \frac{2 \sin^2 x}{\sin x \cos x} = 2 \frac{\sin x}{\cos x} = 2 \tan x \] 
Final Answer: \[ \boxed{2 \tan x} \]