Question

Where A+B=90°. Prove that, (sinB+secA)/sinA=2tanB+tanA.

Answer 1

(sinB+secA)/sinA
= (sinBcosA+1)/SinAcosA
= (cos²A+1)/SinAcosA
= (2- sin²A)/sinACosA
= 2/(sinAcosA) -tanA
= 2(cos²A+sin²A)/(sinAcosA) - tanA
= 2cotA +2tanA-tanA
= 2tanB+TanA