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We have the following formulas for sum of angles when angles are in terms of tan inverse: $\tan^{-1}(x)+\tan^{-1}(y)$ =

  1. $\tan^{-1}((x+y)/(1-xy))$, if $xy<1$

  2. $\pi+\tan^{-1}((x+y)/(1-xy))$, if $x>0,y>0$and $xy>1$

  3. $-\pi +\tan^{-1}((x+y)/(1-xy))$, if $x<0,y<0$ and $xy>1$

I tried it like this : Let $\tan^{-1} (x)=A$ and $\tan^{-1}(y)=B$ where $A,B \in(-(\pi/2),(\pi/2))$

So , $\tan(A+B)=(\tan A+\tan B)/1-\tan A \tan B=(x+y)/1-xy$

$\tan^{-1}((x+y)/(1-xy))=\tan^{-1}(\tan(A+B))$

let $A+B=\alpha$

=$\alpha+\pi$, $-\pi<\alpha<-\pi/2$

$\alpha, -\pi<\alpha <\pi/2$

$\alpha-\pi, \pi/2<\alpha <\pi$

Further how to proceed to get the condition xy><1...?

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    $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Commented Feb 9, 2021 at 7:37

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Well what's happening is, we are doing this step below:

$$\tan^{-1} A + \tan^{-1} B = \tan^{-1} (\tan (\tan^{-1} A + \tan^{-1} B))$$

$\tan^{-1} \tan x$ will shift your input by $n \pi \ ; \ n \ \epsilon \ Z$ to bring it into principal range of tan inverse.

So that step we did above is only valid when you know that $\tan^{-1} A + \tan^{-1} B$ lies within the principal range. To account for other ways are the cases you've listed.

I'm not aware of a purely algebraic approach because I dont get how one would proceed by taking expressions like $\tan^{-1} A + \tan^{-1} B > \frac \pi 2$. Ill present a graphical approach:

enter image description here

That's $y = \tan x$ in $(-\pi, \pi)$

Five cases arise corresponding x,y of opposite signs + to the 4 intervals in x axis, made equally in the above graph when they are of same sign:

0. With $A$ and $B$ of opposite signs, we have $\tan^{-1} A + \tan^{-1}B \ \epsilon \ (-\frac \pi 2, \frac \pi 2)$, which corresponds to the principal range.

1. $\tan^{-1} A + \tan^{-1}B < 0$ and $\tan (\tan^{-1} A + \tan^{-1}B) < 0$: Here since of same sign, both are negative but then since tan is positive you have $1 - AB < 0$

Now $\tan ^{-1} \tan (\tan^{-1} A + \tan^{-1}B)$ gives a value $\pi$ more than what it shoudlve been.

$$\tan^{-1} A + \tan^{-1} B = \tan ^{-1} \left( \frac {A+B}{1-AB} \right) - \pi \ ; \ A,B <0, AB>1$$

Similarly you can do this for the other 3 regions too.

You'll see that regions 0. ($AB<0$), 2.($AB<1; A,B<0$) and 3.($AB<1; A,B>0$) correspond to the same condition of $AB < 1$ and to same result from the expression.

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  • $\begingroup$ What have you taken A and B ? I took Tan ^-1(x)=A and Tan ^-1(y)=B $\endgroup$
    – Rover
    Commented Feb 10, 2021 at 14:14
  • $\begingroup$ Im just taking A and B in $\tan^{−1}A+\tan^{−1}B$ instead of x and y in $\tan^{−1}x+\tan^{−1}y$ so that it doesnt get confusing with x already in use with $y=tan(x)$. I'll edit it to be more consistent $\endgroup$ Commented Feb 10, 2021 at 14:18
  • $\begingroup$ "Here since of same sign, both are negative but then since tan is positive you have 1−xy<0" Can you elaborate.. $\endgroup$
    – Rover
    Commented Feb 10, 2021 at 14:42
  • $\begingroup$ In $\tan \theta$ at region 1. $\theta \ \epsilon \ (-\pi, -\frac \pi2)$ right? Both $\tan ^{-1} A$ and $\tan ^{-1} B$ need to be negative for $\theta = \tan ^{-1}A + \tan ^{-1}B$ to be negative (since they are of same sign). tan inverse negative gives $A,B<0$ too. Which in turn gives $A+B<0$. But at the region we see that $\tan \theta$ is giving a positive value. Thus denominator needs to be negative too. $\endgroup$ Commented Feb 10, 2021 at 14:59
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We know tan has a period of π. So we can write,

$$\tan(\arctan x + \arctan y) = \tan(\arctan x + \arctan y - nπ)\\ \frac{x + y}{1 - xy} = \tan(\arctan x + \arctan y - nπ)\\ \arctan(\frac{x + y}{1 - xy}) + nπ = \arctan x + \arctan y $$

Depending on the RHS, we need to take appropriate value of n.

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