Let $$a = \frac{2 \sin x}{1 + \sin x + \cos x}$$ and $$b = \frac{c}{1 + \sin x}$$. Then a = b, if c = ?

Given that,

$$a = \frac{2 \sin x}{1 + \sin x + \cos x}$$ and $$b = \frac{c}{1 + \sin x}$$

If $$a=b$$

$$ \frac{2 \sin x}{1 + \sin x + \cos x}=\frac{c}{1 + \sin x}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)}{1 + \sin x + \cos x}$$

Now, multiplying and dividing by the conjugate of the denominator, in the given ratio

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(1 + \sin x + \cos x)(1 + \sin x - \cos x)}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(1 + \sin x)^2 -( \cos x)^2}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(1 + \sin^2 x+2\sin x - \cos^2 x)}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(\sin^2 x+\cos^2 x + \sin^2 x+2\sin x - \cos^2 x)}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(2 \sin^2 x+ x+2\sin x) }$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{2\sin x( \sin x+1 )}$$

$$\Rightarrow c= 1 + \sin x - \cos x$$