The (almost really) Complete Works of Lewis Carroll

11530. (Rev. C. L. Dodgson, M.A.)—Required a general investigation of the following trigonometrical formula, which, so far as I know, is new, and very useful in calculating limits for the value of π. The problem, which I set myself, was to break up ${tan}^{-1}1/a$ into two angles of the same form. Let $${tan}^{-1}\frac{1}{a}={tan}^{-1}\frac{1}{a+x}+{tan}^{-1}\frac{1}{a+y}\text{.}$$

Then it is also $$={tan}^{-1}\frac{\frac{1}{a+x}+\frac{1}{a+y}}{1-\frac{1}{a+x}\cdot \frac{1}{a+y}}={tan}^{-1}\frac{2a+x+y}{a^2+a(x+y)+xy-1}\text{.}$$ Then it occurred to me that, if $(xy-1)$ were made equal to $a^2$, the denominator would become $a(2a+x+y)$; i. e., the fraction would become $1/a$. Hence we get the rule: Let $(a^2+1)=xy$; i. e., break up $(a^2+1)$ into any two factors, call them x and y, and use them in the formula with which we began. Thus, if $a=3$, $a^2+1=10=2\times 5$. Hence ${tan}^{-1}\frac{1}{3}={tan}^{-1}\frac{1}{5}+{tan}^{-1}\frac{1}{8}$. By the use of this formula, I have obtained 3⋅141597 and 3⋅141583 as limits for π.