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If cos A, cosB and cosC are the roots o...

If `cos A, cosB and cosC` are the roots of the cubic `x^3 + ax^2 + bx + c = 0` where `A, B, C` are the anglesof a triangle then find the value of `a^2 – 2b– 2c`.

A

0

B

`1//2`

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
C
cos A, cos B and cos C are the roots of the cubic equation
`x^(3)+ax^(2)+bx + c=0`
`rArr cos A+cos B + cos C = -a`
cos A cos B + cos B cos C + cos C cos A = b
cos A cos B cos C = -c
Now `(cos A+cos B cosC)^(2)=(Sigma cos^(2)A)+2(Sigma cos A cos B)`
`therefore cos^(2)A+cos^(2)B+cos^(2)C=a^(2)-2b`
`therefore 3+cos 2A+cos 2B+cos 2C=2a^(2)-4b`
`therefore 3-1-4 cos A cos B cos C = 2a^(2)-4b`
`thererfore 1+2c=a^(2)-2b`
`therefore a^(2)-2b-2c =1`
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