Question

Seca-coseca)(tana+cota+1)=sec^3-cosec^3/seca.Coseca

Answer 1

LHS

=(secA-cosecA)(tanA+cotA+1)

=(1/cosA-1/sinA)(sinA/cosA+cosA/sinA+1)

={(sinA-cosA)/sinAcosA}{(sin²A+cos²A+sinAcosA)/sinAcosA}

=(sinA-cosA)(1+sinAcosA)/(sinAcosA)² [∵, sin²A+cos²A=1]

=(sinA-cosA+sin²AcosA-sinAcos²A)/(sinAcosA)²

=(sinA-sinAcos²A-cosA+sin²AcosA)/(sinAcosA)²

={sinA(1-cos²A)-cosA(1-sin²A)}/(sinAcosA)²

=(sinA.sin²A-cosAcos²A)/(sinAcosA)²

=(sin³A-cos³A)/sin²Acos²A

RHS

=sec³A-cosec³A/secAcosecA

=(1/cos³A-1/sin³A)/(1/cosA×1/sinA)

={(sin³A-cos³A)/sin³Acos³A}/(1/sinAcosA)

=(sin³A-cos³A)/sin³Acos³A×sinAcosA

=(sin³A-cos³A)/sin²Acos²A

∴, LHS=RHS (Proved)