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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