if A+B+C=180° then prove that cos2A + cos 2B - cos 2C = 1-4 sinAsinBcosC
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Answer:
=2.cos(2A+2B)/2.cos (2A-2C)/2-(2cos^2C-1).
= 2.cos(180°-C).cos(A-B)-2cos^2 C. +1.
= - 2.cosC.cos(A-B)-2.cos^2 C. +1
= 1 -2cosC[cos(A-B) +cosC].
= 1 -2.cosC[ cos(A-B)+cos{180°-(A+B)}]
= 1–2cosC.[cos(A-B)-cos(A+B)]
= 1–2.cosC.[2sinA.sinB ]
= 1. - 4.sinA.sinB.cosC. Proved.
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