Question 697167: Please help me verify the identity of cot(θ-(pi)/(2))= -tanθ. I know that -tan0 is not a identity for the equation I just need help figuring out how to find it step by step
please and thank you so very much
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Please help me verify the identity of cot(θ-(pi)/(2))= -tanθ
The best way I know how to verify this is with examples using a unit circle and reference angles.
I will use special angle 30º and degrees in place of radians for for ease of explanation.
First example:
Say, θ=30º in quadrant I where tan>0
The reference angle=30º
cot30º=√3
cot(θ-(pi)/(2)= cot(θ-90º)
This means you are moving θ, 90º clockwise which places the angle is quadrant IV, where cot<0
standard position of the angle=300º
The reference angle is =60º
cot(θ-(pi)/(2)= cot(θ-90º)=cot60º=-√3/3=-tan30º=-tanθ
..
Second example:
Say, θ=210º in quadrant III, where tan>0
The reference angle=30º
cot30º=√3
cot(θ-(pi)/(2)= cot(θ-90º)
This means you are moving θ, 90º clockwise which places the angle is quadrant II, where cot<0
standard position of the angle=120º
The reference angle is =60º
cot(θ-(pi)/(2)= cot(θ-90º)=cot60º=-√3/3=-tan210º=-tanθ
..
Third example:
Say, θ=330º in quadrant IV, where tan<0
The reference angle=30º
cotθ=cot30º=-√3
cot(θ-(pi)/(2)= cot(θ-90º)
This means you are moving θ, 90º clockwise which places the angle is quadrant II, where cot>0
standard position of the angle=240º
The reference angle is =60º
cot(θ-(pi)/(2)= cot(θ-90º)=cot60º=√3/3=-tan330º=-tanθ
..
Note:Only special angle of 30º is used in foregoing examples, but this will work with any angle.
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