R.H.S.
= tanB+tanC
= (sinB/cosB)+(sinC/cosC)
= (sinBcosC+cosBsinC)/(cosB.cosC)
= sin(B+C)/cosA [cosA=cosB.cosC]
= sinA/cosA [A+B+C=pie, B+C=pie-A, sin(B+C)=sin(pie-A)= sinA]
= tanA = L.H.S. Proved.
Q. If A+B+C=pie and cosA=cosB.CosC. Prove that: tanA=tanB+tanC.
R.H.S.
= tanB+tanC
= (sinB/cosB)+(sinC/cosC)
= (sinBcosC+cosBsinC)/(cosB.cosC)
= sin(B+C)/cosA [cosA=cosB.cosC]
= sinA/cosA [A+B+C=pie, B+C=pie-A, sin(B+C)=sin(pie-A)= sinA]
= tanA = L.H.S. Proved.
Q. If A+B+C=pie and cosA=cosB.CosC. Prove that: tanA=tanB+tanC.
S.E.E. Solution
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Soln: R.H.S. = tanB+tanC = (sinB/cosB)+(sinC/cosC) = (sinBcosC+cosBsinC)/(cosB.cosC) = sin(B+C)/cosA [cosA=cosB.cosC]...
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