Q. If A+B+C=pie and cosA=cosB.CosC. Prove that: tanA=tanB+tanC.

Soln:
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.