1
$\begingroup$

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?

$\endgroup$
2
  • 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
2
$\begingroup$

$\dfrac 10=\text{undefined}$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.