Cotangent of Complement equals Tangent

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Theorem

$\map \cot {\dfrac \pi 2 - \theta} = \tan \theta$ for $\theta \ne \paren {2 n + 1} \dfrac \pi 2$

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


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

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


Proof

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


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

From Cosine of Half-Integer Multiple of Pi it follows that this happens when $\theta \ne \paren {2 n + 1} \dfrac \pi 2$.

$\blacksquare$


Also see


Sources

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