In a triangle ABC, right-angled at B, AB+BC=10(1+\sqrt{3}) cm and length of the hypotenuse is 20 cm. What is the value of \tan A+\tan C?

Explanation

Let AB = c and BC = a. Given a+c = 10(1+\sqrt{3}) and a^2+c^2 = 20^2 = 400. Squaring the first equation gives a^2+c^2+2ac = 100(1+3+2\sqrt{3}) = 400+200\sqrt{3}. Substituting a^2+c^2=400, we get 2ac = 200\sqrt{3} \implies ac = 100\sqrt{3}. Now, \tan A + \tan C = \frac{a}{c} + \frac{c}{a} = \frac{a^2+c^2}{ac} = \frac{400}{100\sqrt{3}} = \frac{4}{\sqrt{3}}.

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