Is $\cot(x)=\frac{1}{\tan(x)}$? I just found in some place say that $\cot(x)=\frac{1}{\tan(x)}$. 
If you think about it as a function they are definitely not same. but if you think about as a relation between angle in a right angle triangle and its side they will be equal. 
so the question is its right to say they are equal or not( at least pedagogically)?
Update:-
Even wolfram alpha and Desmos say they are not equal ( have different domain).

The tangent is defined as $$\tan\alpha =\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b}=\frac{\sin~\alpha}{\cos~\alpha}$$
and the cotangent as 
$$\cot\alpha =\frac{\text{adjacent}}{\text{opposite}}=\frac{b}{a}=\frac{\cos\alpha}{\sin\alpha} $$
what leads us to 
$$\frac{\cos\alpha}{\sin\alpha} = \tan\left(\frac{\pi}{2} - \alpha\right) = \frac{1}{\tan \alpha} = \cot\alpha$$
so yes, they are equivalent.
The confusing thing about it is, that you would have to consider the roots of the cosine function, if you were evaluating the tangent function in the denominator before evaluating the fraction.
This is actually a big dissent between different (functional) programming languages, as there is actually no "right order" of evaluation. 
