Question

secA/1-cosecA + cosecA/secA =secA×cosecA

Answer 1

RHS=cosecA+secA)(secAcosecA-2)

=(1/sinA+1/cosA){(1/cosA)(1/sinA)-2}

=[(cosA+sinA)/(sinAcosA) ][(1-2sinAcosA)/(sinAcosA)]

=[(cosA+sinA)/(sinAcosA)][(cosA-sinA)2/(sinAcosA)]

=(cosA+sinA)(cosA-sinA)2/(sin2Acos2A)

=(cosA+sinA)(cosA-sinA)(cosA-sinA)/sin2Acos2A

=cos2A(cosA-sinA)/cos2Asin2A

LHS=(cosecA-secA)(cotA-tanA)

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

=(cosA-sinA/sinAcosA)(cos2A-sin2A)/sinAcosA

=(cosA-sinA)cos2A/sin2Acos2A

Thus LHS=RHS Hence Proved.