Question

A+B+C=90 THEN PT COS 2A + COS 2B + COS 2C = 1 + 4 SINA SINB SINC​

Answer 1

Given A+B+C = 90

A+B = 90 - C

now , we can write

cos2A + cos2B = 2cos(A+B).cos(A-B)  

[use cosA + cosB = 2cos(A+B)/2.cos(A-B)/2 formula]

and

cos2C = 1 - 2sin²C   [use cos2x = 1 - 2sin²x formula]

Substitute these values in the equation

cos2A+cos2B+cos2C

⇒ 2cos(A+B).cos(A-B) + 1 - 2sin²C

⇒2sinC.cos(A-B) + 1 - 2sin²C       [since A+B = 90-C]

⇒ 1 + 2sinC[cos(A-B) - sinC]

⇒1 + 2sinC[cos(A-B) - cos(A+B)]

⇒ 1 + 2sinC(2sinA.sinB)           [since cos(a-b) - cos(a+b) = 2sina.sinb]

⇒ 1 + 4sinA.sinB.sinC = RHS

Hence , proved.

Hope you understand.

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