Question

Please check my reasoning-
For xy>1
\[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{ x+y }{ 1-xy }+\pi\]
But by substituting x=(-a) & y=(-b)
\[\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{ a+b }{ 1-ab }-\pi\]
and still ab>1

Hence, the general formula should be
\[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{ x+y }{ 1-xy }+sgn(x)* \pi\]
for xy>1

Answer 1

PS: I never saw the following formula in books (hence, the need to confirm ;)
\[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{ x+y }{ 1-xy }-\pi\]
for xy>1

Answer 2

so if u dint see it in books did u wrote it by urself ?

Answer 3

I found the (generally accepted) formula to be inconsistent
For xy>1
\[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{ x+y }{ 1-xy }+\pi\]

Answer 4

@eliassaab

Answer 5

but still u need to know
for tan^-1 a = - tan^-1 a
so
\(\large \tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{ x+y }{ 1-xy }+\pi\)
\(\large -\tan^{-1}a-\tan^{-1}b=\tan^{-1}-\frac{(a+b) }{ 1-ab }+\pi\)
\(\large -(\tan^{-1}a+\tan^{-1}b)=-\tan^{-1}\frac{(a+b) }{ 1-ab }+\pi\)
\(\large -(\tan^{-1}a+\tan^{-1}b)=-(\tan^{-1}\frac{(a+b) }{ 1-ab }-\pi)\)
does that help ?!

Answer 6

So, what I posted as question is correct..right ?

Answer 7

yep since xy >1
x sighn = y sighn

Answer 8

Thanxx :)

Answer 9

np :D

Answer 10

Notice that
\[
\tan \left(\tan ^{-1}(x)+\tan
^{-1}(y)\right)=\frac{\tan
\left(\tan ^{-1}(x)\right)+\tan
\left(\tan
^{-1}(x)\right)}{1-\tan
\left(\tan ^{-1}(x)\right) \tan
\left(\tan
^{-1}(x)\right)}=\frac{x+y}{1-x
y}
\\
\tan \left(\tan
^{-1}\left(\frac{x+y}{1-x
y}\right)+\pi \right)=\tan
\left(\tan
^{-1}\left(\frac{x+y}{1-x
y}\right)\right)=\frac{x+y}{1-x
y}
\]
and conclude

Answer 11

Notice that
\[
\tan(u)=\tan(v)\implies u=v+ k \pi \quad k=0,\pm1, \pm2 \pm 3
\]