MCQ
જો $\frac{2 sin \alpha}{1 + sin\alpha + cos \alpha} = \lambda,$ તો $\frac{ 1 + sin \alpha - cos \alpha}{1 + sin\alpha} = .......$
- A$\frac{1}{\lambda}$
- ✓$\lambda$
- C$1 - \lambda$
- D$1 + \lambda$
$\frac{1 + sin \alpha - cos \alpha}{1 + sin \alpha}$
$= \frac{(1 + sin \alpha - cos \alpha)(1+sin \alpha+cos \alpha)}{(1 + sin \alpha) (1 + sin\alpha + cos \alpha)}$
$= \frac{(1 + sin \alpha)^2 - cos^2 \alpha}{(1 + sin\alpha) ((1 + sin \alpha + cos \alpha))}$
$= \frac{2 sin \alpha + 2 sin^2 \alpha }{(1 + sin \alpha) (1 + sin \alpha + cos \alpha)}$
$= \frac{2 sin \alpha}{1 + sin \alpha + cos \alpha}$
$= \lambda$
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