# Explaining the identity $\tan\alpha = \frac{1}{\cot\alpha}$ when $\alpha=0^\circ$ or $\alpha=90^\circ$

According to trigonometry identities , $$\tan\alpha= \frac{1}{\cot\alpha}$$ If $$\alpha = 0^\circ$$ or $$\alpha = 90^\circ$$, place the value of $$\alpha$$, get your result and by observing the result. Give your suggestion on the result.

Result: \begin{align} \tan 0^\circ = \frac{1}{\cot 0^\circ}&\qquad\to\qquad 0 = \frac{1}{\text{not definite}}\\[12pt] \tan 90^\circ = \frac{1}{\cot 90^\circ}&\qquad\to\qquad\text{not definite} =\frac10 \end{align}

Give your suggestion on the result.

• Well, $x=\frac{1}{\frac{1}{x}}$ if and only if $x \neq 0$. – Siminore Apr 1 '14 at 9:36

It is best to see both $\cot$ and $\tan$ through their definitions with $\sin$ and $\cos$: $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha}\\\cot\alpha = \frac{\cos\alpha}{\sin\alpha}$$
It is important to know that $\tan\alpha$ is defined only for values of $\alpha$ for which $\cos\alpha \neq 0$, and $\cot\alpha$ is defined only for values of $\alpha$ for which $\sin\alpha \neq 0$.
Therefore, the equation $$\tan\alpha = \frac{1}{\cot\alpha}$$ is only valid if $\cos\alpha\neq 0$ and $\sin\alpha \neq 0$. For angles $0$ and $90$, that does not hold, so $\tan 0 \neq \frac{1}{\cot 0}$ because $\cot 0$ does not exist.