prove that sinA/1+cosA+1+cosA/sinA=2cosecA
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Answered by
113
here is the answer.
sinA/1+cosA + 1+cosA/sinA
= [sin^2A + (1 + cosA)^2]/sinA(1 + cosA)
= (sin^2A + 1 + cos^2A + 2cosA)/sinA(1 + cosA)
= (1 + 1 + 2cosA)/sinA(1 + cosA) [since,sin^2A + cos^2A = 1]
= (2 + 2cosA)/sinA(1 + cosA)
= 2(1 + cosA)/sinA(1 + cosA) = 2/sinA
= 2 cosecA [since 1/sinA = cosecA]
= R.H.S.
Answered by
128
= sinA/1+cosA + 1+cosA/sinA
= [sin^2A + (1 + cosA)^2]/sinA(1 + cosA)
= (sin^2A + 1 + cos^2A + 2cosA)/sinA(1 + cosA)
= (1 + 1 + 2cosA)/sinA(1 + cosA) [since,sin^2A + cos^2A = 1]
= (2 + 2cosA)/sinA(1 + cosA) = 2(1 + cosA)/sinA(1 + cosA)
= 2/sinA = 2 cosecA [since 1/sinA = cosecA] = R.H.S.
Proved
= [sin^2A + (1 + cosA)^2]/sinA(1 + cosA)
= (sin^2A + 1 + cos^2A + 2cosA)/sinA(1 + cosA)
= (1 + 1 + 2cosA)/sinA(1 + cosA) [since,sin^2A + cos^2A = 1]
= (2 + 2cosA)/sinA(1 + cosA) = 2(1 + cosA)/sinA(1 + cosA)
= 2/sinA = 2 cosecA [since 1/sinA = cosecA] = R.H.S.
Proved
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