Question

Show that in a cyclic quadrilateral ABCD a. tanA + tanB + tanC + tanD =0 b.cos(180-A) + cos(180+B) + cos(180+C) + cos(180+D)

Answer 1

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)
b. cos(180°-A)+cos(180°+B)+cos(180°+C)+cos(180°+D)
=cos{(90°×2)-A}+cos{(90°×2)+B}+cos{(90°×2)+C}+cos{(90°×2)+D}
=-cosA-cosB-cosC-cosD
=-cosA-cosB-cos{(90°×2)-A}-cos{(90°×2)-B}
=-cosA-cosB+cosA+cosB
=0

Answer 2

Answer:

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