If A, B and C are the interior angles of a △ ABC

If A, B and C are the interior angles of a △ ABC, show that

(i) cos (A + B)/2 = sin C/2

(ii) tan (C + A)/2 = cot B/2.

Answer :

A, B and C are the interior angles of a △ ABC

∠A + ∠B + ∠C = 180°

Dividing both sides by 2

(∠A + ∠B + ∠C)/2 = 180°/2

⇒ A/2 + B/2 + C/2 = 90°

(i) cos (A + B)/2 = sin C/2

(A + B)/2 = 90° – C/2

cos (90° – C/2) = sin C/2

Here, cos (90° – θ) = sin θ

sin C/2 = sin C/2

(ii) tan (C + A)/2 = cot B/2

(A + C)/2 = 90° – B/2

= tan (90° – B/2)

= cot B/2

= RHS

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