Question

if A, B, C are interior angles of triangle ABC, prove sinB+C/2.secA/2=1​

Answer 1

Given info : A, B and C are interior angles of triangle ABC.

To prove : sin(B + C)/2 secA/2 = 1

Proof : as it has given that A, B and C are interior angles of triangle ABC.

So, A + B + C = 180°

⇒B + C = 180° - A

⇒(B + C)/2 = (180° - A)/2 = (90° - A/2)

⇒sin(B + C)/2 = sin(90° - A/2)

⇒sin(B + C)/2 = cosA/2....(1) [ as we know, sin(90° - θ) = cosθ]

Now LHS = sin(B + C)/2 secA/2

= cosA/2 secA/2 [ from eq (1) ]

= CosA/2 × 1/cosA/2

= 1 = RHS

Hence proved//