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Question:

In triangle ABC, right-angled at B, if tanA = 1/√3, find the value of: (i) sinAcosC + cosAsinC (ii) cosAcosC - sinAsinC

Solution:

In ΔABC, ∠B = 90°, tanA = 1/√3 = BC/AB
Let BC = x, AB = √3x
AC² = AB² + BC²
AC² = (√3x)² + (x)² = 4x²
AC = 2x
(i) sinAcosC + cosAsinC = (1/2) × (1/2) + (√3/2) × (√3/2) = 1/4 + 3/4 = 1
(ii) cosAcosC - sinAsinC = (√3/2) × (1/2) - (1/2) × (√3/2) = √3/4 - √3/4 = 0