AIME 1984 P13

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Problem Statement

Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$

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Solution

$$\cot^{-1}(x) = \tan^{-1}(\dfrac{1}{x})$$ $$\tan^{-1}(\tan(\cot^{-1}(x))) = \tan^{-1}(\dfrac{1}{x})$$ $$\tan(\cot^{-1}(x)) = \dfrac{1}{x}$$ $$\text{Let } a = \cot^{-1}(3) \implies \tan(a) = \dfrac{1}{3}$$ $$\text{Let } b = \cot^{-1}(7) \implies \tan(b) = \dfrac{1}{7}$$ $$\text{Let } c = \cot^{-1}(13) \implies \tan(c) = \dfrac{1}{13}$$ $$\text{Let } d = \cot^{-1}(21) \implies \tan(d) = \dfrac{1}{21}$$ $$\tan(a + b) = \dfrac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$$ $$\tan(a + b) = \dfrac{\dfrac{1}{3} + \dfrac{1}{7}}{1 - (\dfrac{1}{3})(\dfrac{1}{7})}$$ $$\tan(a + b) = \dfrac{1}{2}$$ $$\tan(c + d) = \dfrac{\tan(c) + \tan(d)}{1 - \tan(c)\tan(d)}$$ $$\tan(c + d) = \dfrac{\dfrac{1}{13} + \dfrac{1}{21}}{1 - (\dfrac{1}{13})(\dfrac{1}{21})}$$ $$\tan(c + d) = \dfrac{1}{8}$$ $$\tan((a + b) + (c + d)) = \dfrac{\tan(a + b) + \tan(c + d)}{1 - \tan(a+b)\tan(c+d)}$$ $$\tan((a + b) + (c + d)) = \dfrac{\dfrac{1}{2} + \dfrac{1}{8}}{1 - (\dfrac{1}{2})(\dfrac{1}{8})}$$ $$\tan((a + b) + (c + d)) = \dfrac{\dfrac{10}{16}}{\dfrac{15}{16}} = \dfrac{2}{3}$$ $$\text{Let } x = 10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21)$$ $$x = 10 \times \dfrac{1}{\tan(a + b + c + d)}$$ $$x = 10 \times \dfrac{1}{\tan((a + b) + (c + d))}$$ $$x = 10 \times \dfrac{1}{\dfrac{2}{3}}$$ $$x = 10 \times \dfrac{3}{2} = 15$$