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Please check the example for clearer understanding of $S_k$

We all know the formula of tan($A_1+A_2+A_3+\cdots+A_n)=$ summation of tan of terms ($A_1,A_2...$) taken once (Let this be $S_1$) - summation of tan of terms taken thrice at a time (Let this be $S_3$) + summation of tan of terms taken 5 at a time (Let this be $S_5$)... (Odd terms at numerator alternatively - and +)
$$÷$$ 1 - summation of tan of terms taken twice at a time (Let this be $S_2$) + summation of tan of terms taken 4 at a time (Let this be $S_4$)... (Even terms at denominator alternatively - and +)

$$\Rightarrow\tan(A_1+A_2+A_3+\cdots+A_n) = \frac{S_1-S_3+S_5-S_7...}{1-S_2+S_4-S_6...}$$
For example lets take $$tan(A+B+C)=\frac{(\tan A+\tan B+\tan C)-(\tan A\tan B\tan C)}{1-(\tan A\tan B+\tan B\tan C+\tan C\tan A)}$$

Here,
$$S_1=\tan A+\tan B+\tan C\\S_2=\tan A\tan B+\tan B\tan C+\tan C\tan A\\S_3=\tan A\tan B\tan C$$

What is the proof for this formula.

Fetray
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2 Answers2

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$$\tan(a + b) ) = \frac{\sin (a+ b)} {\cos(a+b) } = \frac{ \sin a \cos b + \cos a \sin b}{\cos a \cos b - \sin a \sin b} = \frac{ \tan a + \tan b}{ 1-\tan a \tan b}$$

Now recurse by $b \to b + c$ yielding

$$\tan(a + b + c) =\frac{ \tan a + \frac{\tan b +\tan c}{1- \tan b \tan c}}{ 1-\tan a \frac{\tan b +\tan c}{1- \tan b \tan c}} $$ $$= \frac{\tan a + \tan b + \tan c - \tan a \tan b \tan c }{1- \tan a \tan b - \tan a \tan c - \tan b \tan c}$$

that is the quotient

$$\frac{1}{i}\ \frac{ \prod_k (1+ i \tan( a_k)-\prod_k (1- i \tan (a_k))}{ \prod_k (1+ i \tan (a_k))+\prod_k (1- i \tan (a_k))}$$ $$ =\frac{ \Im( \prod_k (1+ i \tan( a_k))}{\Re( \prod_k (1+ i \tan( a_k)))}$$ $$ =\arg \prod_k (1+ i \tan( a_k))$$

Roland F
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  • That last line is a mess, but it is a neat approach to seeing what is happening, you should rewrite it. Start with $1+i\tan x$ having argument $x,$ so $z=\prod (1+i\tan a_i)$ has argument $\sum a_i$ and thus $$\frac{\operatorname{Im} z}{\operatorname{Re} z}=\tan(\sum a_i)$$ – Thomas Andrews May 06 '24 at 17:09
  • You then get that the imaginary part of $z$ is just $S_1-S_3+\cdots$ and the real part is $1-S_2+S_4-\cdots.$ – Thomas Andrews May 06 '24 at 17:12
  • Oh, the problem with this argument is that $1+i\tan x$ does not always have argument $x,$ since it is $\frac{\cos x+i\sin x}{\cos x},$ so the argument is $x$ only when $\cos x>0.$ However, it is always argument $x$ or $x+\pi,$ but $\tan(x+\pi)=\tan x.$ So the quotient of the real part to the imaginary part is $\tan(k\pi+\sum a_i)=\tan(\sum a_i)$ where $k$ is some integer. – Thomas Andrews May 06 '24 at 17:42
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I will expand on the last line of Roland's answer, because it is pretty neat.

Note that the complex number $1+i\tan x=\frac{\cos x+i\sin x}{\cos x}$ has argument $x$ or $x+\pi,$ depending on whether $\cos x$ is positive or negative.

Then $z=(1+i\tan A_1)(1+i\tan A_2)\cdots(1+i\tan A_n)$ has argument $A=A_1+\cdots +A_n+k\pi,$ for some integer $k.$ But $\tan(x+\pi)=\tan(x)$ for all $x,$ so $\tan(A)=\tan(A_1+\cdots +A_n).$

Now you need to note that:

$$z=1+iS_1-S_2-iS_3+S_4+\dots.$$ And $$\tan(A)=\frac{\operatorname{Im} z}{\operatorname{Re}z}=\frac{S_1-S_3+S_5-\cdots}{1-S_2+S_4-\cdots}$$

Thomas Andrews
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