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unknown id · « **Reply #15 on:** March 25, 2008, 02:53:14 pm »

Quote from: AppleXY on March 20, 2008, 08:11:46 am

$\arctan\frac{1}{x} =$ $\frac{\pi}{2}$ - $\arctan{x}$

$tan^{-1} \left(\frac{1}{x} \right) + tan^{-1}(x) = \frac{\pi}{2}$

Since it can be proved that:

$tan^{-1}(x) + tan^{-1}(y) = tan^{-1} \left(\frac{x+y}{1-xy} \right)$ ,

we can use the L.H.S. in the same manner:

$tan^{-1} \left(\frac{1}{x} \right) + tan^{-1}(x)$

$= tan^{-1} \left(\frac{\frac{1}{x}+x}{1-\frac{1}{x} \cdot x} \right)$

$= tan^{-1} \left(\frac{\frac{1}{x}+x}{0} \right)$

$= tan^{-1}($ ∞ $)$

$= \frac{\pi}{2}$

Q.E.D

Mao · « **Reply #16 on:** March 25, 2008, 08:48:42 pm »

Quote from: unknown id on March 25, 2008, 02:53:14 pm

$= tan^{-1} \left(\frac{\frac{1}{x}+x}{0} \right)$

$= tan^{-1}($ ∞ $)$

errr.....
does dividing by 0 really make it $\infty$ ??

brendan · « **Reply #17 on:** March 25, 2008, 08:53:10 pm »

there is something very wrong with your proof unknown id

ed_saifa · « **Reply #18 on:** March 25, 2008, 09:09:41 pm »

lol. It looks like a proof to me.

ed_saifa · « **Reply #19 on:** March 25, 2008, 09:36:07 pm »

$tan^{-1}(x) + tan^{-1}(y) = tan^{-1} \left(\frac{x+y}{1-xy} \right)$ ,

I'm quite lost as to how this was obtained.

unknown id · « **Reply #20 on:** March 25, 2008, 09:46:14 pm »

Quote from: ed_saifa on March 25, 2008, 09:36:07 pm

$tan^{-1}(x) + tan^{-1}(y) = tan^{-1} \left(\frac{x+y}{1-xy} \right)$ ,

I'm quite lost as to how this was obtained.

Let $tan^{-1}(x) = A$ and let $tan^{-1}(y) = B$

Thus,
$tan(A+B) = \left(\frac{tanA + tanB}{1- tanA \cdot tanB} \right)$

$A+B = tan^{-1} \left(\frac{tanA + tanB}{1- tanA \cdot tanB} \right)$

$tan^{-1}(x) + tan^{-1}(y) = tan^{-1} \left(\frac{x + y}{1- xy} \right)$

Q.E.D.

ed_saifa · « **Reply #21 on:** March 25, 2008, 09:57:57 pm »

ah ok. thanks XD

Mao · « **Reply #22 on:** March 25, 2008, 10:03:18 pm »

Quote from: unknown id on March 25, 2008, 09:46:14 pm

$tan^{-1}(x) + tan^{-1}(y) = tan^{-1} \left(\frac{x + y}{1- xy} \right)$

Q.E.D.

but that doesnt make the first one right.....

any explanations?

to answer your question in PM:
you cannot divide by 0, EVER. <-------------fullstop, no buts

there are cases in limits where as the variable approaches 0, the limit reaches infinity (and approaching from the opposite sides can sometimes yield +/- infinity), but NEVER does dividing by zero equal infinity....

NEVER~~~~~~~~~~~~~

AppleXY · « **Reply #23 on:** March 25, 2008, 10:54:25 pm »

Yeah. Unknown id, even though at the surface it looks like a sound proof, it's actually quite flawed. Treating inf and /0 is extremely hypothetical.

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