Question 697167
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 &#952;, 90º clockwise which places the angle is quadrant IV, where cot<0
standard position of the angle=300º
The reference angle is =60º
cot(&#952;-(pi)/(2)= cot(&#952;-90º)=cot60º=-&#8730;3/3=-tan30º=-tan&#952;
..
Second example:
Say, &#952;=210º in quadrant III, where tan>0
The reference angle=30º
cot30º=&#8730;3
cot(&#952;-(pi)/(2)= cot(&#952;-90º)
This means you are moving &#952;, 90º clockwise which places the angle is quadrant II, where cot<0
standard position of the angle=120º
The reference angle is =60º
cot(&#952;-(pi)/(2)= cot(&#952;-90º)=cot60º=-&#8730;3/3=-tan210º=-tan&#952;
..
Third example:
Say, &#952;=330º in quadrant IV, where tan<0
The reference angle=30º
cot&#952;=cot30º=-&#8730;3
cot(&#952;-(pi)/(2)= cot(&#952;-90º)
This means you are moving &#952;, 90º clockwise which places the angle is quadrant II, where cot>0
standard position of the angle=240º
The reference angle is =60º
cot(&#952;-(pi)/(2)= cot(&#952;-90º)=cot60º=&#8730;3/3=-tan330º=-tan&#952;
..
Note:Only special angle of 30º is used in foregoing examples, but this will work with any angle.