Question:
$\frac{(2 \sin \alpha)(1 + \sin \alpha)}{(1 + \sin \alpha + \cos \alpha)(1 + \sin \alpha - \cos \alpha)} =$
$\frac{(2 \sin \alpha)(1 + \sin \alpha)}{(1 + \sin \alpha + \cos \alpha)(1 + \sin \alpha - \cos \alpha)} =$
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Grouping $(1+\sin \alpha)$ together helps simplify the denominator quickly using the difference of squares.
Updated On: Apr 28, 2026
- $\tan \alpha$
- $\frac{\sin \alpha + 1}{\sin \alpha - 1}$
- 1
- 2
- $\frac{\cos \alpha + 1}{\cos \alpha - 1}$
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The Correct Option is C
Solution and Explanation
Step 1: Analysis
Simplify the denominator using the identity $(a+b)(a-b) = a^2 - b^2$: $[(1 + \sin \alpha) + \cos \alpha][(1 + \sin \alpha) - \cos \alpha] = (1 + \sin \alpha)^2 - \cos^2 \alpha$.
Step 2: Calculation
$(1 + \sin^2 \alpha + 2 \sin \alpha) - \cos^2 \alpha$. Since $1 - \cos^2 \alpha = \sin^2 \alpha$, the denominator becomes $\sin^2 \alpha + \sin^2 \alpha + 2 \sin \alpha = 2 \sin^2 \alpha + 2 \sin \alpha$.
Step 3: Conclusion
Denominator $= 2 \sin \alpha (1 + \sin \alpha)$. Since the numerator is also $2 \sin \alpha (1 + \sin \alpha)$, the ratio is 1. Final Answer: (C)
Simplify the denominator using the identity $(a+b)(a-b) = a^2 - b^2$: $[(1 + \sin \alpha) + \cos \alpha][(1 + \sin \alpha) - \cos \alpha] = (1 + \sin \alpha)^2 - \cos^2 \alpha$.
Step 2: Calculation
$(1 + \sin^2 \alpha + 2 \sin \alpha) - \cos^2 \alpha$. Since $1 - \cos^2 \alpha = \sin^2 \alpha$, the denominator becomes $\sin^2 \alpha + \sin^2 \alpha + 2 \sin \alpha = 2 \sin^2 \alpha + 2 \sin \alpha$.
Step 3: Conclusion
Denominator $= 2 \sin \alpha (1 + \sin \alpha)$. Since the numerator is also $2 \sin \alpha (1 + \sin \alpha)$, the ratio is 1. Final Answer: (C)
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