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प्रश्न
In an ∆ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C
उत्तर
L.H.S = a cos A+ 6 cos B + c cos C
Using since formula, we get k sin A cos A + k sin B cos B + k sin C cos C k
= `"k"/2` [2 sin A cos A + 2 sin B cos B + 2 sin C cos C]
= `"k"/2` [sin 2A + sin 2B + sin 2C]
= `"k"/2` [2 sin(A + B) . cos(A – B) + 2 sin C . cos C]
= `"k"/2` [2 sin(A – B) . cos(A – B) + 2 sin C . cos C]
= `"k"/2` [2 sin C . cos(A – B) + 2 sin C . cos C]
= k sin C [cos(A – B) + cos C]
= `"k" sin "C" [cos("A" - "B") + cos (pi - bar("A" + "B"))]`
= k sin C [cos(A – B) – cos(A + B)]
= k sin C . 2 sin A sin B
= 2k sin A . sin B sin C
= 2a sin B sin C
= R.H.S
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