Cotangent of Complement equals Tangent/Corollary 2

Corollary to Cotangent of Complement equals Tangent

Let $x \ne \paren {4 n + 1} \dfrac \pi 4$

Then:

$\ds \map \cot {\dfrac \pi 4 - x } = \map \tan {\dfrac \pi 4 + x}$

where:

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

Proof

\(\ds \map \cot {\dfrac \pi 2 - \theta}\)

\(=\)

\(\ds \tan \theta\)

Cotangent of Complement equals Tangent

\(\ds \leadsto \ \ \)

\(\ds \map \cot {\dfrac \pi 2 - \paren {\dfrac \pi 4 + x } }\)

\(=\)

\(\ds \map \tan {\dfrac \pi 4 + x}\)

let $\theta = \dfrac \pi 4 + x$

\(\ds \leadsto \ \ \)

\(\ds \map \cot {\dfrac \pi 4 - x }\)

\(=\)

\(\ds \map \tan {\dfrac \pi 4 + x}\)

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

Since $\theta \ne \paren {2 n + 1} \dfrac \pi 2$. and $\theta = \dfrac \pi 4 + x$

Then $x \ne \paren {4 n + 1} \dfrac \pi 4$

$\blacksquare$