Question

True or false? tan( pi/2 -x)=cotx

Answer 1

True.

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.

`\tan((\pi)/(2)-x)`

Rewrite using quotient identity for tangent

`(\sin((\pi)/(2)-x))/(\cos((\pi)/(2)-x))`

Rewrite using difference identities for sine and cosine

`(\sin((\pi)/(2))\cos(x)-\sin(x)\cos((\pi)/(2)))/(\cos((\pi)/(2))\cos(x)+\sin((\pi)/(2))\sin(x))`

sin(pi/2)=1 while cos(pi/2)=0

`(1 \cdot \cos(x)-\sin(x) \cdot 0)/(0 \cdot \cos(x)+1 \cdot \sin(x))`

Do a little basic algebra

`(\cos(x)-0)/(0+\sin(x))`

More simplification

`(\cos(x))/(\sin(x))`

This is quotient identity for cotangent

`\cot(x)`

Answer 2

True

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)

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1 sin(x) +0 cos(x)

tan (pi/2 -x) = cos (x)

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1 sin(x)

We know cos(x)/ sin (x) = cot(x)

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