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Evaluate ; $\cos 90^\circ \times \tan 90^\circ$

$$\cos 90^\circ \times \tan 90^\circ$$ $$ 0 \times \infty$$ $$0$$.

Or,

$$\cos 90 \times \dfrac {\sin 90}{\cos 90}$$ $$\sin 90$$ $$1$$

Or,

$$0 \times \dfrac {1}{0}$$ $$\dfrac {0}{0}$$ $$\infty$$

Which one is correct? Or, are there other alternatives?

  • 0
    It's futile to talk about $\tan 90^\circ$ though you can evaluate $\lim_{x\to \pi/2}\tan x\cos x$(which is $1$). – 2017-02-12

3 Answers 3

0

We cannot cancel it like this actually, cancelling is possible only due to the multiplicative inverse that is for example $2.\frac{1}{2} = 1$ because $\frac{1}{2}$ is the multiplicative inverse of $2$ , also $0$ has no multiplicative inverse here , now in your second argument you cannot cancel $\cos(90)$ term because $\cos(90) = 0$ has no multiplicative inverse here.

3

For cancellation you have to ensure that you cannot divide by zero.

3

$\infty$ is not a number, so you cannot use it in operations. Also, $\tan(90°)$ is not equal to $\infty$, but simply does not exist. So you cannot multiply it by anything.

This means that $\cos(90°)\times\tan(90°)$ in undefined, that is does not exist.