Question

of sinA = a/b, show that sec A + tanA = √a+b/a-b​

Answer 1

Answer:

ya so?

Step-by-step explanation:

Answer 2

Step-by-step explanation:

sinA=opposite side/hypotenuse

tanA=opp.side/adj.side

tanA=a/(b^2-a^2)^1/2

secA=hypotenuse/adj.side

secA=b/(b^2-a^2)^1/2

secA+tanA=a+b/(b^2-a^2)^1/2

=a+b/(a+b) ^1/2(b-a)^1/2

=(a+b/a-b) ^1/2