Question 22

$$\frac{(2 \sin A)(1 + \sin A)}{1 + \sin A + \cos A}$$ is equal to:

Solution

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


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