Question

15. In AABC, the value of cos (A-B)-cos (C) is​

Answer 1

Answer:

cos(A-B) - cosC = 2cosBcosC

Explanation:

As wkt, sum of three angles of triangle = 180°

i.e., A+B+C=180°

Now, cos(A-B) - cosC = 2 sin(A-B+C)/2 sin(C-A+B)/2

= 2 sin(180°-B-B)/2 sin(180°-A-A)/2

= 2 sin(180°-2B)/2 sin(180°-2A)/2

= 2 sin(90°-B) sin(90°-A)

= 2cosBcosC