Question 54

What is the value of $$\frac{[\sin (90 - A) + \cos (180 - 2A)]}{[\cos (90 - 2A) + \sin (180 - A)]}$$?

Solution

$$\frac{[\sin(90-A)+\cos(180-2A)]}{[\cos(90-2A)+\sin(180-A)]}$$

$$=\frac{[\cos A-\cos2A]}{[\sin(2A)+\sin(A)]}\ .$$

$$=\frac{2.\sin\frac{3A}{2}.\ \sin\frac{A}{2}}{2.\ \sin\frac{3A}{2}.\cos\frac{A}{2}}\ .$$

$$=\tan\frac{A}{2}\ .$$

C is correct choice.

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