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Pi formula proofs

Proof of Leibniz's formula

Machin's formula

\frac{π}{4} = 4 \tan^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239}

Recall the formulas:

\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}

\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}

\tan (2 x) = \frac{2 \tan x}{1 - \tan^{2} x}

Let

\tan α = \frac{1}{5}

We can obtain tan(2α) = 5/12 and tan(4α) = 120/119 by using the above formula. Therefore,

\tan (4 \tan^{- 1} \frac{1}{5}) = \frac{120}{119}

Consider,

\tan (4 \tan^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239}) = \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \frac{1}{239}} = \frac{120 \cdot 239 - 119}{119 \cdot 239 + 120} = \frac{120 (120 + 119) - 119}{119 (120 + 119) + 120} = \frac{12 0^{2} + 11 9^{2}}{12 0^{2} + 11 9^{2}} = 1

4 \tan^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239} = \frac{π}{4}