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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

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3 years ago

```							Solving RHS2cosAcosBcosC =(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^2Bso,1 - 2cosAcosBcosC = 1 + cos^2C + cos^2A - sin^2B= 1 - sin^2B + cos^2C + cos^2A= cos^2B + cos^2C + cos^2A
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2 years 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^2Bso,1 - 2cosAcosBcosC = 1 + cos^2C + cos^2A - sin^2B= 1 - sin^2B + cos^2C + cos^2A= cos^2B + cos^2C + cos^2AThanks and Regards
```
3 months ago
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