Question

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

• A. \( \frac{1-\sin x}{\cos x} \)
• B. \( \frac{1+\sin x}{\cos x} \)
• C. \( \frac{1+\cos x}{\sin x} \)
• D. \( \frac{1-\cos x}{\sin x} \)

Hint:

When \(\tan x\) and \(\sec x\) are present together, convert them into \(\sin x\) and \(\cos x\) to simplify easily.

Answer 1

The Correct Option is B

Concept: To simplify expressions containing \(\tan x\) and \(\sec x\), convert them into \(\sin x\) and \(\cos x\).

Step 1: Given expression: \[ \frac{\tan x-1+\sec x}{\tan x-\sec x+1} \] Use: \[ \tan x=\frac{\sin x}{\cos x}, \qquad \sec x=\frac{1}{\cos x} \]

Step 2: Substitute these values. \[ \frac{\frac{\sin x}{\cos x}-1+\frac{1}{\cos x}} {\frac{\sin x}{\cos x}-\frac{1}{\cos x}+1} \]

Step 3: Simplify numerator. \[ \frac{\sin x+1-\cos x}{\cos x} \]

Step 4: Simplify denominator. \[ \frac{\sin x-1+\cos x}{\cos x} \] So the expression becomes: \[ \frac{\sin x+1-\cos x}{\sin x-1+\cos x} \]

Step 5: The simplified standard form is: \[ \frac{1+\sin x}{\cos x} \] Therefore, \[ \boxed{\frac{1+\sin x}{\cos x}} \]