arctan(cot10Pi/9)

Answer 1

x = arctan[cot(10pi/9)]
= arctan[1/tan(10pi/9)
= arctan[1/tan((9+1)pi/9)]
= arctan{1/tan([(9/9)+(1/9)]pi)}
= arctan[1/tan(pi/9)]

tan(x)-1/tan(pi/9)=0
[tan(pi/2)*tan(x)-1]/tan(pi/2)=0
tan(pi/2)*tan(x)-1=0

Express in sin and cos. Use identities also. (lol, go solve the details. too lazy to do it.)

--> cos[(pi/9)+x]=0

Find the arcos of both sides in radian form.

(pi/9)+x = pi/2
x = (pi/2)- (pi/9)
x = 7pi/18

Answer 2

arctan(cot(x))
= arctan(tan(pi/2-x))
= pi/2-x

This is not quite true when dealing with QIII angles...