Question

$\frac{(2 \sin \alpha)(1 + \sin \alpha)}{(1 + \sin \alpha + \cos \alpha)(1 + \sin \alpha - \cos \alpha)} =$

• A. $\tan \alpha$
• B. $\frac{\sin \alpha + 1}{\sin \alpha - 1}$
• C. 1
• D. 2
• E. $\frac{\cos \alpha + 1}{\cos \alpha - 1}$

Hint:

Grouping $(1+\sin \alpha)$ together helps simplify the denominator quickly using the difference of squares.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are asked to simplify a complex trigonometric fraction. A key strategy here is to look for algebraic patterns, such as the difference of squares, to simplify the denominator.
Step 2: Key Formula or Approach:
1. Algebraic identity (Difference of Squares): \( (A+B)(A-B) = A^2 - B^2 \)
2. Pythagorean identity: \( \sin^2 \alpha + \cos^2 \alpha = 1 \), which implies \( 1 - \cos^2 \alpha = \sin^2 \alpha \).
Step 3: Detailed Explanation:
Let's focus on simplifying the denominator first:
\[ (1+\sin\alpha+\cos\alpha)(1+\sin\alpha-\cos\alpha) \] We can group the terms to apply the difference of squares formula. Let \( A = (1+\sin\alpha) \) and \( B = \cos\alpha \). The expression is in the form \( (A+B)(A-B) \).
\[ (A+B)(A-B) = A^2 - B^2 = (1+\sin\alpha)^2 - (\cos\alpha)^2 \] Expand \( (1+\sin\alpha)^2 \):
\[ (1 + 2\sin\alpha + \sin^2\alpha) - \cos^2\alpha \] Now, use the Pythagorean identity to replace \( \cos^2\alpha \) with \( 1 - \sin^2\alpha \):
\[ 1 + 2\sin\alpha + \sin^2\alpha - (1 - \sin^2\alpha) \] Distribute the negative sign:
\[ 1 + 2\sin\alpha + \sin^2\alpha - 1 + \sin^2\alpha \] Combine like terms:
\[ (1-1) + 2\sin\alpha + (\sin^2\alpha + \sin^2\alpha) = 2\sin\alpha + 2\sin^2\alpha \] Factor out the common term \( 2\sin\alpha \):
\[ 2\sin\alpha(1+\sin\alpha) \] Now, substitute this simplified denominator back into the original fraction:
\[ \frac{(2\sin\alpha)(1+\sin\alpha)}{2\sin\alpha(1+\sin\alpha)} \] Assuming \( \sin\alpha \neq 0 \) and \( \sin\alpha \neq -1 \), we can cancel the entire numerator and denominator.
\[ = 1 \] Step 4: Final Answer:
The value of the expression is 1.