Question

If 3 cot A = 2, then find the value of `(4sin"A" - 3cos"A")/(2sin"A" + 3cos"A")`

Answer 1

`cot A = "adjacent"/"opposite" = 2/3`

• Adjacent side = 2

• Opposite side = 3

• Hypotenuse = `sqrt(2^2 +3^2) = sqrt13`

sin A = `3/sqrt13`

cos A = `2/sqrt13`

`(4 sin A - 3 cos A)/(2 sin A + 3 cos A) = (4(3/sqrt13) - 3(2/sqrt13))/(2(3/sqrt13) + 3(2/sqrt13))`

`(6/sqrt13)/(12/sqrt13) = 6/12`

`=1/2`