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Question 9 (ii)
In triangle ABC, right-angled at B, if tan A = 13, 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 = 13
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) cosAcosCsinAsinC=(32×12)(12×32)=3434=0


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