Why Does tan(x + pi/2) Equal -cotx in Trigonometry?

[mathdad]

I decided to review a little trigonometry.

Why does tan(x + pi/2) = -cotx?

I cannot use the tangent of a sum formula because
tan(pi/2) does not exist.

How about tan(x + pi/2) = [sin(x + pi/2)]/[cos(x + pi/2)] and then apply the addition rules for sine and cosine?

 

[MarkFL]

You could certainly do that, however, you could also write:

\(\displaystyle \tan\left(x+\frac{\pi}{2}\right)=\tan\left(\frac{\pi}{2}-(-x)\right)\)

Using a co-function identity, we obtain:

\(\displaystyle \tan\left(x+\frac{\pi}{2}\right)=\cot\left(-x\right)\)

Using the fact that the cotangent function is odd, we have:

\(\displaystyle \tan\left(x+\frac{\pi}{2}\right)=-\cot\left(x\right)\) :D

You can also use the angle-sum identity for tangent, if you then use a limit and L'Hôpital's Rule for the resulting indeterminate form, however we'll keep this pre-calc. ;)

 

[mathdad]

You know the short cut way. Cool.