proving $\tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$

Question

(1) How ca we prove $\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$

(2) How ca we prove $\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = -\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$when $x<0,y<0,xy>1$

$\bf{My\; Try::}$ My Try for $(1)\;,$ Let $\tan^{-1}(x)=A\Rightarrow x = \tan (A)$ and $\tan^{-1}(y) = B\Rightarrow y = \tan(B)$

Now $\displaystyle \tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\cdot \tan B} = \frac{x+y}{1-xy}$

and Given $x>0\Rightarrow \tan A>0$ and $y>0\Rightarrow \tan B>0$ and given $xy>1$

So $\displaystyle \frac{x+y}{1-xy}<0\Rightarrow \tan (A+B)<0$

Means $(A+B)$ lies in $\bf(II)$ quadrant or in $\bf{(IV)}$ quadrant.

But above $\displaystyle \tan (A)\;\;,\tan (B)>0\Rightarrow 0<A,B<\frac{\pi}{2}$.

So $\displaystyle 0<A+B<\pi$(Means $(A+B)$ lies on $\bf{I}$ quadrant.)

Now I did not understand how can i solve it.

Help me

Thanks

@chubakueno by that logic you can prove that $1=-1$ by squaring both sides... You cannot assume what you are trying to prove. — E.O., May 15 '14 at 03:18
@E.O. I am aware, every move has to be properly justified. juantheron is doing fine, because he is doing exactly that. I do not understand what does he doesnt understand, though. — chubakueno, May 15 '14 at 03:23
http://math.stackexchange.com/questions/326334/a-question-about-the-arctangent-addition-formula — lab bhattacharjee, May 15 '14 at 03:51

John Graham · Answer 1 · 2023-02-27T16:39:14.790

It is quite simple ; lets consider tan(θ) = x and let it be so that tan(ϕ) = y , then we know , θ = tan⁻¹(x) and ϕ = tan⁻¹(y)

$\displaystyle\Rightarrow$ tan(θ+ϕ)= $\displaystyle(\frac{tanθ+tanϕ}{1-tanθ*tanϕ})\;$

$\displaystyle\Rightarrow$ tan(θ+ϕ) = $\displaystyle(\frac{x+y}{1-x*y})\;$ as [Note : tan(θ) = x and tan(ϕ) = y]

$\displaystyle\Rightarrow$ $\displaystyle \pi $ is an interval of tan , which means tan(a-n*$\pi$) is same as tan(a) where $n∈ℤ$

using that :

$\displaystyle\Rightarrow$ tan(θ+ϕ-$\displaystyle \pi)$ = $\displaystyle(\frac{x+y}{1-x*y})\;$

$\displaystyle\Rightarrow$ θ+ϕ-$\displaystyle \pi$ = tan⁻¹$\displaystyle(\frac{x+y}{1-x*y})\;$

and alas simply :

∴$\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$

conversly using n=-1 , we can observe ,

$\displaystyle\Rightarrow$ tan(θ+ϕ+$\displaystyle \pi)$ = $\displaystyle(\frac{x+y}{1-x*y})\;$

$\displaystyle\Rightarrow$ θ+ϕ+$\displaystyle \pi$ = tan⁻¹$\displaystyle(\frac{x+y}{1-x*y})\;$

hence,

∴$\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = -\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$when $x<0,y<0,xy>1$

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