If \[ a=\frac{2\sin x}{1+\cos x+\sin x} \] Prove that \[ \frac{1-\cos x+\sin x}{1+\sin x}=a \]

Solution:

\[ a=\frac{2\sin x}{1+\cos x+\sin x} \]

Multiply numerator and denominator by \[ 1-\cos x+\sin x \]

\[ a= \frac{2\sin x(1-\cos x+\sin x)} {(1+\cos x+\sin x)(1-\cos x+\sin x)} \]

\[ = \frac{2\sin x(1-\cos x+\sin x)} {(1+\sin x)^2-\cos^2 x} \]

\[ = \frac{2\sin x(1-\cos x+\sin x)} {1+2\sin x+\sin^2 x-\cos^2 x} \]

Using \[ \sin^2 x+\cos^2 x=1 \]

\[ = \frac{2\sin x(1-\cos x+\sin x)} {2\sin x(1+\sin x)} \]

\[ = \frac{1-\cos x+\sin x} {1+\sin x} \]

Hence proved.