SOLUTION: Evaluate the cot[arccos(5-x)]

Question 713908: Evaluate the cot[arccos(5-x)]
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Evaluate the cot[arccos(5-x)]
let A=adjacent side
let O=opposite side
H=hypotenuse
..
cot[arccos(5-x)]
This reads: cot of an angle whose cos=(5-x)
cos(5-x)=A/H=(5-x)/1
A=(5-x)
H=1
O=√(H^2-A^2) (Pythagorean Theorem)
=√[1^2-(5-x)^2]
=√[1-(25-10x+x^2)]
O=√(-24+10x-x^2)
cot[arccos(5-x)]=A/O=(5-x)/√(-24+10x-x^2)
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