math problem...very very hard!!!?

Question

Find Sec(arcSin(x-3)/(x+2)) show me the work

Answer 1

Let arcSin((x-3)/(x+2)) = k

Therefore sin(k) = (x-3)/(x+2)

sec(x) = 1/cos(x)

cos(x) = sqrt(1 - sin^2(x))

Therefore, sec(x) = 1/sqrt(1 - sin^2(x))

since sin(k) = (x-3)/(x+2), we get,

cos(k) = sqrt(1-((x-3)/(x+2))^2)

Therefore sec(k) = 1/sqrt(1-((x-3)/(x+2))^2)

Simplifying the above expression we get,

sec(k) = (x+2)/sqrt(10x - 5)

ie sec(arcSin((x-3)/(x+2))) = (x+2)/sqrt(10x - 5)

Answer 2

Let's break this down from the inside out. If I could make a drawing things would be much simpler. Consider first arcsin(x-3/x+2). This function can be thought of as representing the angle at the vertex of a triangle with an opposite side length of x-3 and a hypotenuse of x+2. Using the Pythagorean theorem, we can show the adjacent side has a length of SQR(10x-5)

With this triangle drawn, we can now state that the secant of this angle is

(x+2)/SQR(10x-5)

Answer 3

(x-3) / (x+2)

((x-3) / (x+2)) (x+2)

(x-3)(x+2)/(x+2)

(x-3)

Answer 4

The answer would just be (x+2)/(x-3).