Consider the LHS of the expression.
⇒cosA1+sinA+1+sinAcosA
⇒cos2A+(1+sinA)2cosA(1+sinA)
⇒sin2A+cos2A+1+2sinAcosA(1+sinA)
⇒2+2sinAcosA(1+sinA)
⇒2(1+sinA)cosA(1+sinA)
⇒2cosA
⇒2secA=RHS
Hence, proved.