given that

Asked by  | 10th Sep, 2012, 10:57: PM

Expert Answer:

(1+cosA)(1+cosB)(1+cosC) = (1-cosA)(1-cosB)(1-cosC)
Dividing both sides with (1-cosa)(1-cosb)(1-cosc), we get:
 
 
(1+cosa)(1+cosb)(1+cosc) / (1-cosa)(1-cosb)(1-cosc) = 1
 
Multiplying the numerator and the denominator by (1-cosa)(1-cosb)(1-cosc), we get:
 
{(1+cosa)(1-cosa)(1+cosb)(1-cosb)(1+cosc)(1-cosc)} / {(1-cosa)(1-cosa)(1-cosb)(1-cosb)(1-cosc)(1-cosc)}
 
Using the identity sin^2 x + coa^2 x = 1 and taking square root on both sides, we get
(sina.sinb.sinc) = (1-cosa)(1-cosb)(1-cosc)
 
Thus, we have:
(1+cosA)(1+cosB)(1+cosC) = (1-cosA)(1-cosB)(1-cosC) = sina sinb sinc
 
Hence the result.

Answered by  | 11th Sep, 2012, 09:46: AM

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