Question

A, B, C and D are the four angles of A cyclic quadrilateral taken in order, prove that

i) Tan A + Tan B + Tan C + Tan D = 0​

Answer 1

Given

A, B, C and D are the four angles of a cyclic quadrilateral.

To Prove

tanA + tanB + tanC + tanD = 0

Proof

As written they are in cyclic quadrilateral. So,

“Sum of opposite angle is 180°.”

Taking L.H.S.

⇒ ∠A + ∠C = 180°

⇒ ∠A = 180° - ∠C

⇒ ∠B + ∠D = 180°

⇒ ∠B = 180° - ∠D

Now,

⇒ tanA = - tanC

⇒ tanA + tanC = 0

Similarly,

⇒ tanB = - tanD

⇒ tanB + tanD = 0

Now,

⇒ tanA + tanB + tanC + tanD

⇒ (tanA + tanC) + (tanB + tanD)

⇒ 0 + 0

⇒ 0

L.H.S. = R.H.S.

Hence, proved

Answer 2

Step-by-step explanation:

Since A, B, C and D are the angles of cyclic quadrilateral.

/_A+/_C=180° and /_B+/_D=180°( sum of opposite sides of a qyadrilateral is 180°).

therefore, tanA+tanB+tanC+tanD

=tan(180°-tanC)+tan(180°-tanD)+tanC+tanD

=tan{(90°×2)-C}+tan{(90°×2)-D}+tanC+tanD

= -tanC-tanD+tanC+tanD

=0

hence, proved.

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hope It helps you!!

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