Question

proof if tan^-1x + tan^-1y + tan^-1z= pi/2
then xy +yz+zx=1

Answer 1

this is what i done so far
\[\tan ^{-1} \left( \frac{ x+y }{ 1-xy } \right) + \tan ^{-1}z =\pi/2\]

Answer 2

\[\frac{ x+y }{ 1-xy } +z = \tan \pi/2\]

Answer 3

let
a = arctan(x)
b = arctan(y)
c= arctan(z)

then
x = tan a
y = tan b
z = tan c


xy + yz + zx = (tan a)*tan(b) + (tan b ) * tan c + tan c * tan a

Answer 4

forgot to mention i need to use
\[\tan ^{-1}x + \tan ^{-1}y= \tan ^{-1}\left( \frac{ x+y }{ 1-xy } \right)\]

Answer 5

Let \(\tan^{-1}x=\alpha, \tan^{-1}y = \beta, \tan^{-1}y = \gamma\).\[\alpha + \beta + \gamma =\pi/2\]\[\tan (\alpha + \beta + \gamma) = \frac{\tan \alpha + \tan \beta + \tan \gamma - \tan \alpha \tan \beta \tan \gamma}{ 1 - \tan \alpha \tan \beta - \tan \beta \tan \gamma - \tan \gamma \tan \alpha }\]\[= \frac{x+y+z-xyz}{1-(xy+yz+zx)}=\tan \pi/2\]In order to be undefined, \(xy+yz+zx = 1\).

Answer 6

from what you've done so far
\(\tan ^{-1} \left( \frac{ x+y }{ 1-xy } \right) + \tan ^{-1}z =\pi/2\)

\(\tan ^{-1} \left( \frac{ x+y }{ 1-xy } \right) =\pi/2 - \tan ^{-1}z\)

then take tans on both sides....

Answer 7

@IrishBoy123 done

Answer 8

You can memorise the formula:\[\tan (x_1 + x_2 + x_3 + \cdots) = \frac{S_1 - S_3 + S_5 - S_7 + \cdots}{S_0 - S_2 + S_4 - S_6 + \cdots}\]Where \(S_n\) is the cyclic sum of tangents taken \(n\) at a time.

Answer 9

Apply the given information twice, you can get the answer
\(tan^{-1}x + tan^{-1}y= tan^{-1}\dfrac{x+y}{1-xy}\)
now, \(\dfrac{x+y}{1-xy}is X, ~and~z~is~Y\)
You have \(tan^{-1}X+tan^{-1}Y= tan^{-1}\dfrac{X+Y}{1-XY}=\dfrac{\pi}{2}\)
That is \(tan (\pi/2) = \dfrac{X+Y}{1-XY}\)
and we know that tan (pi/2) is undefined, hence \(1-XY=0\implies XY=1\)
As above, \(X =\dfrac{x+y}{1-xy}, Y =z\), hence \(XY = \dfrac{x+y}{1-xy}z=1\)
that is (x+y)z = 1-xy, hence xz +yz +xy =1

Answer 10

@Loser66, when pi/2 is undefined, what happen to the numerator, X+Y?

Answer 11

a fraction is undefined when the denominator =0 no matter what the numerator is. So that we don't care about the numerator.

Answer 12

Note: pi/2 is not undefined, but tan (pi/2) is undefined.

Answer 13

thank you so much, i finally understand

Answer 14

$$ \large {
\tan ^{-1} \left( \frac{ x+y }{ 1-xy } \right) + \tan ^{-1}z =\pi/2

\\ \tan ^{-1} \left( \frac{ x+y }{ 1-xy } \right) =\pi/2 - \tan ^{-1}z
\\ \tan(\tan ^{-1} \left( \frac{ x+y }{ 1-xy } \right) )=\tan(\pi/2 - \tan ^{-1}z)
\\ \frac{ x+y }{ 1-xy }=\cot( \tan ^{-1}z)
\\ \frac{ x+y }{ 1-xy } =1 / (\tan( \tan ^{-1}z))
\\ \frac{ x+y }{ 1-xy } =1 / z
}$$

Answer 15

@jayzdd, thanks :)

Answer 16

credit goes to Parthkholi + irishboy

Answer 17

thank you so much guys, could not done without you

Answer 18

@jayzdd why there is cot??