Used Formula:

sin 2θ = 2 sin θ cos θ

cos C + cos D = 2cos((C + D)/2) cos((C − D)/2)

Calculation:

sin A − cos B − cos C = 0

⇒ sin A = cos B + cos C

Using the formula given above

⇒ 2 sin(A/2) cos(A/2) = 2cos((B + C)/2) cos((B − C)/2)

We know that, for ΔABC
∠A + ∠B + ∠C = π

⇒ sin(A/2) cos(A/2) = cos((π − A)/2) cos((B − C)/2)

Since, cos(π/2 − θ) = sin θ

⇒ sin(A/2) cos(A/2) = sin(A/2) cos((B − C)/2)

⇒ cos(A/2) = cos((B − C)/2)

⇒ A/2 = (B − C)/2

⇒ A = B − C

⇒ B = A + C

But ∠A + ∠B + ∠C = π

⇒ B = π − B

⇒ 2B = π

⇒ B = π/2

∴ Angle B is equal to π/2.