Since $\tan x=\sin x/\cos x$ and $\cot x=1/\tan x$, can we redefine cotangent as $\cot x =\cos x/\sin x$? if we use this definition, we can find this value $\cot \pi/2= 0$. What are the advantages of the second definition?
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1You might find it interesting that $\tan$ and $\cot$ are the only trigonometric functions that are both reciprocals, $\cot x = 1 / \tan x$, and complements, $\cot x = \tan(\pi/2 - x)$, hence the name. – cr3 Dec 18 '15 at 19:02
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I would say that $$\cot x=\frac{\cos x}{\sin x}$$ is "the" definition. The equation $$\cot x=\frac1{\tan x}$$ is rather an identity--that is, it is an equation that is true whenever both sides are defined. This fails to be a proper definition because as you say, this doesn't tell us (for example) what $\cot\frac\pi2$ is. It is very nearly a definition, though--it allows us to infer the values of $\cot x$ when $x$ is an odd multiple of $\frac\pi2$ by continuity. Alternately, if we use the convention that $\frac1{-\infty}=\frac1{+\infty}=0,$ then it is a definition--however, since we usually think of trigonometric functions as real-valued (and $\pm\infty$ are not real numbers), this is not desirable.
Cameron Buie
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Yeah, that's an identity.
One advantage of this identity is if you're asked to simplify something like this:
$$\cot(x)\sec(x)$$
$$=\left(\frac{\cos x}{\sin x}\right)\left(\frac{1}{\cos x}\right)$$
$$=\frac{1}{\sin x}$$
$$=\csc x$$
$$\require{cancel}$$
Leonidas Lanier
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