The Student Room Group

Reply 1

1/tanx = cot x

tan^-1x is the inverse of tan. This is more formally known as arctan. Its a completely different to 1/tanx.

Reply 2

tan 63.4 = 2, so tan^-1( 2) = 63.4

Meanwhile 1 / tan 63.4 = 1/2 = 0.5

Don't know how else to explain it....

Reply 3

tan^-1 is an inverse function, defined such that tan^-1(tan(x)) = tan(tan^-1(x)) = x, for most 'useful' values of x. It is true that, say, a^-1 = 1/a, but "tan" conforms to function notation and not to standard algebraic notation. It's a shame the two overlap, but it is very standard to define a function f^-1 such that f^-1(f(x)) = f(f^-1(x)) = x. arctan is another way of writing tan^-1, but obviously you can't write 'arcf' for the inverse of f, so tan^-1 stuck for convenience.

1/tanx isn't a function. It's just 1 divided by tanx.

Reply 4

generalebriety

1/tanx isn't a function.

well...

Reply 5

chewwy
well...

Blah.

Reply 6

generalebriety

1/tanx isn't a function. It's just 1 divided by tanx.


wtf lol

Reply 7

Unparseable latex formula:

[br]\frac{1}{tan(\theta)}\ =\ cot(\theta)\\[br]tan^{-1}(\theta)\ = arctanx[br]


Where arctan(θ)arctan(\theta) is the inverse function of tan(θ)tan(\theta)

Reply 8

I'm tempted to have another rant about notation...
But I'll resist. Suffice to say:

If you mean INVERSE TAN, write ARCTAN
If you mean 1/TAN, write COT

Reply 9

jamesgurung
I'm tempted to have another rant about notation...
But I'll resist. Suffice to say:

If you mean INVERSE TAN, write ARCTAN
If you mean 1/TAN, write COT

Fighting a losing battle here. Besides, I don't see the problem with tan-1x = arctanx, given that it follows standard function notation. You wouldn't dream of thinking f-1(x) = 1/f(x), would you? And you certainly wouldn't write arcf. And if you meant 1/f, you wouldn't call it something entirely different. It's a notational inconvenience, but tan-1 is perfectly unambiguous, given that no one uses it to mean 1/tan - they'd just write cot.

Reply 10

You make a good point. Very logical. Remember that f^-1 does have the ambiguity of representing the 'preimage of f', though. :wink: