Question

If $\frac{2 sin \alpha }{1 + cos ⁡ \alpha + sin ⁡ \alpha }=\frac{3}{4}$ , then the value of $\frac{ 1 - cos \alpha + sin ⁡ \alpha }{1 + sin ⁡ \alpha }$ is equal to

• A. $4/3$
• B. $3/4$
• C. $1/4$
• D. $7/4$

Answer 1

Answer: (B)

Let, $x=\frac{2 sin \alpha }{1 + cos ⁡ \alpha + sin ⁡ \alpha }$
Let, $y=\frac{1 - cos \alpha + sin ⁡ \alpha }{1 + sin ⁡ \alpha }.$
Now, $\frac{x}{y}=\frac{2 sin \alpha \left(1 + sin ⁡ \alpha \right)}{\left(1 + sin ⁡ \alpha \right)^{2} - \left(cos\right)^{2} ⁡ \alpha }$
$\Rightarrow \frac{x}{y}=\frac{2 sin \alpha \left(1 + sin ⁡ \alpha \right)}{1 + \left(sin\right)^{2} ⁡ \alpha + 2 sin ⁡ \alpha - 1 + \left(sin\right)^{2} ⁡ \alpha }=1$
$\Rightarrow x=y=3/4$