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Solution
$$\frac{(2 \sin A)(1 + \sin A)}{1 + \sin A + \cos A}$$
=Â $$\frac{(2 \sin A + 2\sin^2 A)}{1 + \sin A + \cos A}$$
= $$\frac{(2 \sin A + 2 -Â 2\cos^2 A)}{1 + \sin A + \cos A}$$
($$\because \sin^2 A +Â \cos^2 A = 1$$)
=Â $$\frac{(2 \sin A + 1 +Â \sin^2 A + \cos^2 A - 2\cos^2 A)}{1 + \sin A + \cos A}$$
= $$\frac{((\sin A + 1)^2 - \cos^2 A)}{1 + \sin A + \cos A}$$
($$\because (a)^2 - (b)^2 = (a +Â b)(a - b)$$)
=Â $$\frac{(1 + \sin A + \cos A)(1 + \sin A - \cos A)}{1 + \sin A + \cos A}$$
=Â $$(1 + \sin A - \cos A)$$
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