Answer:
True.
Step-by-step explanation:
Let's use the picture I made.
I used degrees instead...
tan(90-x)= b/a . I did opposite over adjacent for the angle labeled 90-x which is that angle's measurement.
cot(x)=b/a . I did adjacent over opposite for the angle labeled 90 which is that angle's measurement.
Now this is also known as a co-function identity.
[tex]\tan(\frac{\pi}{2}-x)[/tex]
Rewrite using quotient identity for tangent
[tex]\frac{\sin(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x)}[/tex]
Rewrite using difference identities for sine and cosine
[tex]\frac{\sin(\frac{\pi}{2})\cos(x)-\sin(x)\cos(\frac{\pi}{2})}{\cos(\frac{\pi}{2})\cos(x)+\sin(\frac{\pi}{2})\sin(x)}[/tex]
sin(pi/2)=1 while cos(pi/2)=0
[tex]\frac{1 \cdot \cos(x)-\sin(x) \cdot 0}{0 \cdot \cos(x)+1 \cdot \sin(x)}[/tex]
Do a little basic algebra
[tex]\frac{\cos(x)-0}{0+\sin(x)}[/tex]
More simplification
[tex]\frac{\cos(x)}{\sin(x)}[/tex]
This is quotient identity for cotangent
[tex]\cot(x)[/tex]
Answer:
True
Step-by-step explanation:
tan( pi/2 -x)
We know that tan (a-b) = sin (a-b) / cos (a-b)
tan (pi/2 -x) = sin (pi/2 -x)
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cos (pi/2 -x)
We know that
sin (a-b) = sin(a) cos(b) - cos(a) sin(b)
and cos (a-b) = sin(a) sin(b) + cos(a) cos(b)
tan (pi/2 -x) = sin (pi/2) cos (x) - cos (pi/2) sin (x)
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sin(pi/2) sin(x) + cos(pi/2) cos(x)
We know sin (pi/2)=1
cos (pi/2) = 0
tan (pi/2 -x) = 1 cos (x) - 0 sin (x)
----------------------------------------------
1 sin(x) +0 cos(x)
tan (pi/2 -x) = cos (x)
------------------
1 sin(x)
We know cos(x)/ sin (x) = cot(x)
tan (pi/2 -x) = cot(x)
