Tangent of Complement equals Cotangent

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Theorem

$\map \tan {\dfrac \pi 2 - \theta} = \cot \theta$ for $\theta \ne n \pi$

where $\tan$ and $\cot$ are tangent and cotangent respectively.


That is, the cotangent of an angle is the tangent of its complement.

This relation is defined wherever $\sin \theta \ne 0$.


Proof

\(\ds \map \tan {\frac \pi 2 - \theta}\) \(=\) \(\ds \frac {\map \sin {\frac \pi 2 - \theta} } {\map \cos {\frac \pi 2 - \theta} }\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \frac {\cos \theta} {\sin \theta}\) Sine and Cosine of Complementary Angles
\(\ds \) \(=\) \(\ds \cot \theta\) Cotangent is Cosine divided by Sine


The above is valid only where $\sin \theta \ne 0$, as otherwise $\dfrac {\cos \theta} {\sin \theta}$ is undefined.

From Sine of Multiple of Pi it follows that this happens when $\theta \ne n \pi$.

$\blacksquare$


Also see


Sources

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