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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