How do you find the exact value of # cot(arcsin ((-7/13))#?

1 Answer
Jun 17, 2018

#+-2sqrt(30)/7#

Explanation:

it depends on how you define range of #arcsin(theta)#
1. range of #arcsin(theta) = {y|0 < y <= pi/2 , pi < y <=3/2*pi}#
The angle of #arcsin(-7/13)# is at third quadrant.
#cot(arcsin(-7/13))=-sqrt(13^2-(-7)^2)/(-7)=2sqrt(30)/7#
2. range of #arcsin(theta) = {y|-pi/2 <= y <0, 0 < y <= pi/2}#
The angle of #arcsin(-7/13)# is at fourth quadrant.
#cot(arcsin(-7/13))=sqrt(13^2-(-7)^2)/(-7)=-2sqrt(30)/7#