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if cosA+sinA=secA+cosecA=n, prove that n(m*m-1)=2m. |
n(m*m-1)=(secA+cosecA)*((cosA+sinA)*(cosA+sinA)-1) = (secA+cosecA)*((cos^2A+sinAcosA+sinAcosA+sin^2A)-1) = (secA+cosecA)*((1+2sinAcosA)-1) = ((secA+cosecA)*(1+2sinAcosA-1)) =(secA+cosecA)*(2sinAcosA) =2secAsinAcosA + 2cosecAsinAcosA =2sinA+2CosA =2m |
