Question

secA+tanA=3/2 then sin A

Answer 1

Hi !

Given,

SecA + tanA = 3/2

1/cosA + sinA/cosA = 3/2

sinA+1/cosA = 3/2

Squaring both sides,we get ,

(sinA +1)²/cos²A = 9/4

[cos²A can be written as 1 - sin²A]

(sinA +1) (sinA +1) / 1 - sin²A = 9/4


the denominator is of the form (a+b)(a-b) = a² - b²

(sinA +1) (sinA +1) / (1 - sinA)(1+ sinA) = 9/4

sinA + 1 /  1 - sinA = 9/4

cross multiplication,

4(sinA + 1) = 9( 1 - sinA)

4sinA + 4 = 9 - 9 sinA


13 sinA = 5

sinA = 5/13