Question : If $x=\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}$, then the value of $\frac{1-\cos \theta+\sin \theta}{(1+\sin \theta)}$ is:
Option 1: $\frac{x}{(1+x)}$
Option 2: $x$
Option 3: $\frac{1}{x}$
Option 4: $\frac{(1+x)}{x}$
Correct Answer: $x$
Solution :
Given expression,
$x=\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}$
Multiply and divide by $1-\cos \theta+\sin \theta$,
⇒ $x=\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}\times\frac{1-\cos \theta+\sin \theta}{1-\cos \theta+\sin \theta}$
⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\cos \theta+\sin \theta)({1-\cos \theta+\sin \theta})}$
⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)^2-(\cos^2\theta)}$
⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)^2-(1-\sin^2\theta)}$ [As $\cos^2\theta+\sin^2\theta=1$]
⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)^2-(1-\sin\theta)(1+\sin\theta)}$
⇒ $=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)[(1+\sin\theta)-(1-\sin\theta)]}$
⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)(2\sin\theta)}$
⇒ $\frac{(1-\cos \theta+\sin \theta)}{(1+\sin\theta)}=x$
Hence, the correct answer is $x$.
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