Question

Prove: (secA-cosecA)(tanA+cotA+1) = sec 3 A - cosec 3 A/ 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)