trigonometry - Proving that $\tan \theta=\cot(90^\circ-\theta)$ when $\theta>90^\circ$ - Mathematics Stack Exchange

I'm asked to prove $\tan \theta=\cot (90-\theta)$ So $\tan \theta =\frac{a}{b}=\cot(90^\circ-\theta)$ But what if $\theta>90?$

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If [math] extrm{cosec } heta = rac{p+q}{p-q}[/math] then [math

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tan(90 – x), tan(90 – A), tan(90 – theta)

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