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
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