Question

Question 9 (ii) In triangle ABC, right-angled at B, if tan A = 1√3, find the value of: (ii) cos Acos C - sinAsin C

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Solution

Let ΔABC in which ∠B=90∘, According to the question, Tan A = BCAB = 1√3 Let AB = √3k and BC = k, where k is a positive real number. By Pythagoras theorem in ΔABC; we get, AC2=AB2+BC2 AC2=(√3k)2+(k)2 AC2=3k2+k2 AC2=4k2 AC=2k sin A = BCAC = 12 sin C = ABAC =√32 cos A = ABAC = √32 cos C = BCAC = 12 (ii) cosAcosC−sinAsinC=(√32×12)−(12×√32)=√34−√34=0

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