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Question
Expand cos(A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B, if A + B + C = `pi/2`
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Solution
Taking A + B = X and C = Y
We get cos(X + Y) = cos X cos Y – sin X sin Y
(i.e) cos(A + B + C) = cos(A + B) cos C – sin(A + B) sin C
= (cos A cos B – sin A sin B) cos C – [sin A cos B + cos A sin B] sin C
cos(A + B + C) = cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C
If (A + B + C) = `π/2` then cos(A + B + C) = 0
⇒ cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C = 0
⇒ cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin
C sin A cos B
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Choose the correct alternative:
`(sin("A" - "B"))/(cos"A" cos"B") + (sin("B" - "C"))/(cos"B" cos"C") + (sin("C" - "A"))/(cos"C" cos"A")` is
