If A ,B ,C ,D are the angles of a cyclic quadrilateral, then prove that cotA + cotB + cotC+ cotD=0

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Solution

$G i v e n A, B C, D a r e t h e a n g l e s i n c y l i c q u a d r i l a t e r a l t h e n ∠ A + ∠ C = 180^{0} ∠ B + ∠ D = 180^{0} c o t A + c o t B + c o t C + c o t D = c o t A + c o t B + c o t (180 - A) + c o t (180 - B) = c o t A + c o t B - c o t A - c o t B (∴ c o t (180 - x) = - c o t x) = 0$

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