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if cos(β+α), cosβ , cos(β-α) are in H.P. then show that cosβ.secα/2 = +-√2

if cos(β+α), cosβ , cos(β-α) are in H.P. then show that cosβ.secα/2 = +-√2

Grade:11

1 Answers

Arun
25763 Points
3 years ago
Dear Abhishek
 
 cos(a-b),cosa,cos(a+b) are in HP, then1/cos(a-b), 1/cosa , 1/cos(a+b) are in A.P.    1/cos(a+b) – 1/cosa = 1/cosa – 1/cos(a-b)    [cosa – cos(a+b)] / cos(a+b).cosa = [cos(a-b) – cosa] / cos(a-b).cosa    cos(a-b)[cosa – cos(a+b)] = cos(a+b)[cos(a-b) – cosa]    cos(a-b).cosa – [cos(a-b).cos(a+b)] = [cos(a-b).cos(a+b)] – cos(a+b).cosa    cos(a-b).cosa + cos(a+b).cosa = 2.cos(a-b).cos(a+b)    cosa [cos(a-b) + cos(a+b)] = 2(cos²a – sin²b)    cosa.(2.cosa.cosb) = 2(cos²a – sin²b)    cos²a.cosb – cos²a = -sin²b    cos²a.(cosb – 1) = -(1 – cos²b)    cos²a.(cosb – 1) = (cosb – 1)(cosb + 1)    cos²a – 1 = cosbCos B cos A/2 = +- sqrt (2) 

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