Question

If $\frac{(2\sin \alpha )}{(1+\cos \alpha +\sin \alpha )}=x$, then $\frac{(1-\cos \alpha +\sin \alpha )}{(1+\sin \alpha )}$ is equal to

• A. $\frac{1}{x}$
• B. $x$
• C. $1-x$
• D. $1+x$

Answer 1

The correct option is B

$x$

Explanation for the correct option:

Step 1. Find the value of given expression:

Given, $\frac{(2\sin \alpha )}{(1+\cos \alpha +\sin \alpha )}=x$ …..(1)

Now, $\frac{(1-\cos \alpha +\sin \alpha )}{(1+\sin \alpha )}$

Step 2. by rationalizing the numerator, we get

$=\frac{(1-\cos \alpha +\sin \alpha )}{(1+\sin \alpha )}\times \frac{\left(1+\sin \alpha +\cos \alpha \right)}{\left(1+\sin \alpha +\cos \alpha \right)}$

$=\frac{{(1+\sin \alpha )}^2-{\cos }^2\alpha }{(1+\sin \alpha )\left(1+\sin \alpha +\cos \alpha \right)}$ $\left[\mathbf{\because }{\mathbf{(}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{a}}^2\mathbf{}\mathbf{+}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathit{a}\mathit{b}\right]$

$=\frac{1+{\sin }^2\alpha +2\sin \alpha -\left(1-{\sin }^2\alpha \right)}{(1+\sin \alpha )\left(1+\sin \alpha +\cos \alpha \right)}$ $\left[\mathbf{\because }{\mathbf{sin}}^{\mathbf{2}}\mathit{x}\mathbf{+}{\mathbf{cos}}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathbf{1}\right]$

$=\frac{2{\sin }^2\alpha +2\sin \alpha }{(1+\sin \alpha )\left(1+\sin \alpha +\cos \alpha \right)}$

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

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

$=\mathit{x}$ $\left[\because Fromequation1\right]$

Hence, Option $B$ is Correct.