Why are the tangents for 90 and 270 degrees undefined?

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Why are the tangents for 90 and 270 degrees undefined?

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Answer 1 · 2015-10-15T18:21:53

This is a good question indeed!

Explanation:

I'll try to give you a visual explanation.
Consider the trigonometric meaning of tangent of an angle $α$ $alpha$ :
enter image source here
The tangent of $α$ $alpha$ is equal to the length of the segment $A B$ $AB$ ; but when $α$ $alpha$ becomes $90^{\circ}$ $90^@$ the length of $A B$ $AB$ get stretched upwards (or downwards for $α = 270^{\circ}$ $alpha=270^@$ ) so that we will never meet $A$ $A$ !!!
enter image source here

Hope it helps!

Answer 2 · 2015-10-15T20:10:31

You can also say that $tan x = \frac{sin x}{cos x}$ $tanx = sinx/cosx$ .

You can take a look at the overlap of $sin x$ $sinx$ and $cos x$ $cosx$ below to see what happens to each as we approach $90^{o}$ $90^o$ and $270^{o}$ $270^o$ from the right or left:

graph{(y - sinx)(y - cosx) = 0 [-0.034, 6.2831, -1.2, 1.2]}

We can establish that:

$lim_(x->90^(o^(-))) sinx/cosx = -1/0 = -oo$

because $cosx$ decreases while $sinx$ increases as $x->90^o$ from the left, and

$lim_(x->90^(o^(+))) sinx/cosx = 1/0 = oo$

because both $cosx$ and $sinx$ increase as $x->90^o$ from the right.

We can also see that:

$lim_(x->270^(o^(-))) sinx/cosx = -1/0 = -oo$

because $sinx$ decreases but $cosx$ increases as $x->270^o$ from the left, and

$lim_(x->270^(o^(+))) sinx/cosx = 1/0 = oo$

because both $sinx$ and $cosx$ decrease as $x->270^o$ from the right.

Since the limits from the left and right side are not the same, $\mathbf(tanx)$ of those angles is undefined.

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Why are the tangents for 90 and 270 degrees undefined?

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