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tan A + tan B+tan C+tanD0100t cos (180 + B) |
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Answer» Since A,B,C and D are the angles of a cyclic quadrilateral, then∠A+∠C=180° and∠B+∠D=180°∴, a.tanA+tanB+tanC+tanD=tan(180°-C)+tan(180°-D)+tanC+tanD=tan{(90°×2)-C}+tan{(90°×2)-D}+tanC+tanD=-tanC-tanD+tanC+tanD=0 (Proved) |
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