Prove that:
cosA1+sinA+1+sinAcosA=2SecA
To prove:
Taking LHS
cosA1+sinA+1+sinAcosA
=(cosA)(cosA)+(1+sinA)(1+sinA)(1+sinA)(cosA) ( Taking LCM of denominator)
=cos2A+(1+sinA)2(1+sinA)cosA=cos2A+1+sin2A+2sinA(1+sinA)cosA
=2+2sinA(1+sinA)cosA (sin2A+cos2A=1)
=2(1+sinA)(1+sinA)cosA=2cosA=2SecA
=RHS
Hence, proved cosA1+sinA+1+sinAcosA=2SecA
Prove the identity cos2A/1+sinA+1+sinA/cosA=2secA
Prove that cosA1+sinA+1+sinAcosA=2secA