Question

\((1+\tan A+\sec A)(1+\sec A-\tan A)-2\sec A=\)

• A. \(0\)
• B. \(-2\)
• C. \(1\)
• D. \(2\)

Hint:

For expressions containing \((1+\sec A+\tan A)(1+\sec A-\tan A)\), use: \[ (a+b)(a-b)=a^2-b^2 \] and: \[ \sec^2 A-\tan^2 A=1 \]

Answer 1

The Correct Option is D

Concept:
This question is based on algebraic identity and trigonometric identity. We will use: \[ (a+b)(a-b)=a^2-b^2 \] and: \[ \sec^2 A-\tan^2 A=1 \]

Step 1: Rewrite the expression.
Given expression: \[ (1+\tan A+\sec A)(1+\sec A-\tan A)-2\sec A \] Rearrange the first factor: \[ 1+\tan A+\sec A=1+\sec A+\tan A \] Second factor: \[ 1+\sec A-\tan A \] Let: \[ x=1+\sec A \] and: \[ y=\tan A \] Then the product becomes: \[ (x+y)(x-y) \]

Step 2: Apply identity.
Using: \[ (x+y)(x-y)=x^2-y^2 \] we get: \[ (1+\sec A+\tan A)(1+\sec A-\tan A) \] \[ =(1+\sec A)^2-\tan^2 A \] So the full expression becomes: \[ (1+\sec A)^2-\tan^2 A-2\sec A \]

Step 3: Expand the square.
\[ (1+\sec A)^2=1+2\sec A+\sec^2 A \] So: \[ 1+2\sec A+\sec^2 A-\tan^2 A-2\sec A \]

Step 4: Cancel like terms.
\[ +2\sec A-2\sec A=0 \] So expression becomes: \[ 1+\sec^2 A-\tan^2 A \]

Step 5: Use trigonometric identity.
We know: \[ \sec^2 A-\tan^2 A=1 \] Therefore: \[ 1+\sec^2 A-\tan^2 A=1+1 \] \[ =2 \]

Step 6: Check the options.
Option (A) \(0\) is incorrect.
Option (B) \(-2\) is incorrect.
Option (C) \(1\) is incorrect.
Option (D) \(2\) is correct. Hence, the correct answer is: \[ \boxed{(D)\ 2} \]