In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of sin A cos C + cos A sin C.

Given that: tan A = 1/√3
Let tan A = 1k/√3, where k is a real number.
In △ABC, by Pythagoras theorem, we have
AC² = BC²+AB²
= (1k)²+(V3k)²
= k²+3k² = 4k²
⇒ AC = √(4k)² = 2k
(i) sin A cos C+cos A sin C
= 1/2 × 1/2 + √3/2 × √3/2 = 1/4 + 3/4 = (1+3)/4 = 4/4 = 1
(ii) cos A cos C sin A sin C
= √3/2 × 1/2 – 1/2 × √3/2 = √3/4 – √3/4 = 0

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