As you all know $$\tan(x)=\frac{\cos(x)}{\sin(x)}$$so, $$\cos(x)\cdot \tan(x)=\sin(x)$$ $\sin\left(\dfrac{\pi}{2}\right)=1$ and $\cos\left(\dfrac{\pi}{2}\right)=0$ also $\tan\left(\dfrac{\pi}{2}\right)=\text{undefined}$

However $\tan\left(\dfrac{\pi}{2}\right)$ is not equal but is undefined because it is $\dfrac{1}{0}$ so does that mean $\dfrac{1}{0} \cdot 0=1$ in this case?

  • 3
    $\begingroup$ No. The first statement is only true when the denominator is nonzero. You cannot divide by $0$. Ever. $\endgroup$ – saulspatz Mar 14 '18 at 20:58
  • $\begingroup$ It's important to keep in mind that a trig identity (or any identity) is true ONLY WHEN the constituent parts are defined. Consequently, the relation $\cos x \cdot \tan x = \sin x$ is NOT assumed to hold at $x = \pi/2$. $\endgroup$ – Blue Mar 14 '18 at 23:27

$\dfrac 10=\text{undefined}$

And you can never divide by $0$ and/or work with undefined numbers (why even call them numbers if they are undefined).

In this case, $0$ and $0$ do not "cancel each other out".

If you can argue that $\dfrac 10\cdot0=1$, then another person can pose an argument that $\dfrac10\cdot0=0$.

Therefore, it is senseless to define such an expression. $\tan\left(\dfrac{\pi}2\right)$ is undefined in mathematical terms.


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