Question

If ABCD is a cyclic quadrilateral and A,B,C,D are its interior angles , then prove that

tanA/2+tanB/2=cotC/2+cotD/2

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Answer 1

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Step-by-step explanation:

In a cyclic quadrilateral, opposite angles are supplementary. This means ...

A + C = 180° ⇒ A/2 +C/2 = 90° ⇒ C/2 = 90° -A/2

B + D = 180° ⇒ B/2 +D/2 = 90° ⇒ D/2 = 90° -B/2

It is a trig identity that ...

tan(α) = cot(90° -α)

so we have ...

tan(A/2) = cot(90° -A/2) = cot(C/2)

and

tan(B/2) = cot(90° -B/2) = cot(D/2)

Adding these two equations together gives the desired result:

tan(A/2) +tan(B/2) = cot(C/2) +cot(D/2)