Question

Prove that: (secA+cosecA)(sinA+cosA)=tanA+cotA+2=2+secAcosecA​

Answer 1

(sinA+cosA)(secA+cosecA)

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

={(sinA+CosA)(sinA+cosA)}/{sinA×cosA}

=(sinA+cosA)²/(sinA×cosA)

=(sin²A+cos²A+2sinAcosA)/(sinA×cosA)

=(1+2sinAcosA)/(sinA×cosA)

=2+secAcosecA

LHS = RHS

hence proved

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