(secA-cosA)(cosecA-sinA)=tanA/1+tan^2A

Question

i want the answer

Answer 1

(1/cosA-cosA)(1/sinA-sinA)=(1-cos^2A)(1-sin^2A)/cosAsinA=sin^2Acos^2A/cosAsinA=sinAcosA.

tanA/(1+tan^2A)=sinA/cosA*1/sec^2A=sinA/cosA*cos^2A=sinAcosA.

Since both expressions simplify to sinAcosA, (secA-cosA)(cosecA-sinA)=tanA/(1+tan^2A).

Answer 2

LHS = (secA-cosA)(cosecA-sinA)

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

        = {(1-cos^2A)/cosA}{(1-sin^2A)/sinA}

        = sin^2Acos^2A/sinAcosA

        = sinAcosA

        = {sinA/cosA}cos^2A

        = tanA/sec^2A

        = tanA/(1+tan^2A)