Question: "Can I put x=π2 in tan⁡2x=2tan⁡x1−tan2x?"

Chris Cruz

Answered question

2022-01-14

Can I put

x = \frac{π}{2}

\tan 2 x = \frac{2 \tan x}{1 - \tan^{2} x}

kaluitagf

Beginner2022-01-19Added 38 answers

Actually x,2x both should be

\equiv\equiv \frac{π}{2} (mod π)

to keep

\tan x, \tan 2 x

defined
But we can take limit and set

\tan x = \frac{1}{t}

to find

L = lim_{x \to \frac{π}{2}} \frac{2 \tan x}{1 - \tan^{2} x} = lim_{t \to 0} \frac{\frac{2}{t}}{1 - \frac{1}{t^{2}}} = lim_{t \to 0} \frac{2 t}{t^{2} - 1} as t \neq 0 as t \to 0

\Rightarrow L = 0

RizerMix

Expert2022-01-20Added 656 answers

Notice,

\tan x

is undefined at

x = n π + \frac{π}{2}

. Similarly

\tan 2 x

is undefined at

2 x = n π + \frac{π}{2}

Therefore for defining

\tan x

\tan 2 x

x \neq n π + \frac{π}{2}

2 x \neq n π + \frac{π}{2}

alenahelenash

Expert2022-01-24Added 556 answers

x = \frac{π}{2} \pm ϵ

with

ϵ > 0, \tan x \sim \mp \frac{1}{ϵ}

, so

\tan \frac{π}{2}

is undefined. (It's not even

\infty

, because it approaches

\mp \infty

depending on the direction.) But now

\tan 2 x \sim \frac{\mp \frac{2}{ϵ}}{- O (\frac{1}{ϵ^{2}})} \sin \pm 2 ϵ

which has

ϵ \to 0

limit 0. This agrees with

\tan π = 0

, of course.

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