Question

A+B=90,prove that (secA+sinB)/sinA=tanA+2tanB

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.....mark as brainliest