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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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)

she will most likely have 33  

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