SinB+SecA/sinA=2tanB +tanA
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Answer:
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
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