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If cos square A + cos square b + cos square C + 2 cos a × cos b × Cosc= 1 if a + b + C = 180 degree

If cos square A + cos square b + cos square C + 2 cos a × cos b × Cosc= 1 if a + b + C = 180 degree

Question Image
Grade:11

2 Answers

repalle kishore babu
16 Points
2 years ago
Solving RHS
2cosAcosBcosC =(2cosAcosB)cosC
=(cos(A+B) + cos(A-B))cosC
=cos(pi-C)cosC + cos(A-B)cos(pi-(A+B))
=-2cos^2C - 2cos(A+B)cos(A-B)
=-cos^2C - cos^2A + sin^2B

so,

1 - 2cosAcosBcosC = 1 + cos^2C + cos^2A - sin^2B
= 1 - sin^2B + cos^2C + cos^2A
= cos^2B + cos^2C + cos^2A
Rishi Sharma
askIITians Faculty 646 Points
10 months ago
Dear Student,
Please find below the solution to your problem.

2cosAcosBcosC =(2cosAcosB)cosC
=(cos(A+B) + cos(A-B))cosC
=cos(pi-C)cosC + cos(A-B)cos(pi-(A+B))
=-2cos^2C - 2cos(A+B)cos(A-B)
=-cos^2C - cos^2A + sin^2B
so,
1 - 2cosAcosBcosC = 1 + cos^2C + cos^2A - sin^2B
= 1 - sin^2B + cos^2C + cos^2A
= cos^2B + cos^2C + cos^2A

Thanks and Regards

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