SOME TRIG FUNTION SOLVING PROBLEM

I have a few question about solving trig function. It's near final exam, but seems like i foget all the things. =[.. Can someone explain them in detail?

1. evaluate sin(tan-1(9/15))

2. cotΘ=-4/3 secΘ=?

THANX A LOT!!!!!

1)

sin [tan^-1(9/15)]

let tan^-1(9/15) = A

tan A = 9/15 = 3/5

opposite side = 3 and adjacent side = 5, hypotenuse = √34

sin A = 3/√34

sin [tan^-1(9/15)] = sin A = 3/√34

2)

cot Θ = - 4/3

adjacent side = 4 and opposite side = 3, hypotenuse = √(4^2 + 3^2) = 5

sec Θ = hypotenuse/adj.side = - 5/4 ( if Θ is in II quadrant) or 5/4 (if Θ is in IV quadrant )

1) tan-1(9/15) finds an angle the tan of which is 9/15 - you can consider this the opposite of the tan function just divide is the opposite of multiply. So find the angle and then find the sin of this angle.

You can also work this out using pythagoras getting the answer 9/sqrt(306).

The answer is 0.514495...

2 Use cot =1/tan and sec = 1/cos

This gives cos/sin = -4cos/3

So sinΘ = -3/4

so Θ = -48 degrees or thereabouts

1. hypotenuse = sqrt(9^2+15^2) = sqrt(306)
sin(tan-1(9/15)) = 9/sqrt(306) = 9sqrt(306)/306= sqrt(306)/34

2. hypotenuse = 5
secΘ=-5/4

cot x = - 4/3

cos x / sin x = -4/3

cos x = -4 sin x / 3

1/cos x = sec x = -3 / (4 sin x)






sin (tan^(-1) (9/15) = sin [(tan)^(-1) (3/5)]


= sin ( 30.96)

= 0.5145