Trigonometric EquationHard

Question

In a triangle ABC with fixed base BC, the vertex A moves such that cosB + cosC = 4sin2 . If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, then

Options

A.b + c = 4a
B.b + c = 2a
C.locus of point A is an ellipse
D.locus of point A is a pair of straight lines

Solution

2 cos cos = 4 sin2
cos = 2 sin (A/2)
⇒  
⇒  
⇒   b + c = 2a (constant).

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