Question

in ∆abc the value of cos(a-b)-cos(c) is​

Answer 1

Answer:

2cosacosb

Step-by-step explanation:

cos(a-b)-cosc

=-2sin(a-b+c)/2*sin(a-b-c)/2

=-2sin(180-b-b)*sin{a-(b+c)]/2

-2sin(90-b)*sin(a-(180-a))/2

=-2cosb*sin(a-180+a)/2

=-cosb*sin(2a-180)/2

=-2cosb*sin{-(90-a)}

=2cosb*sin(90-a)

=2cosacosb

Answer 2

Answer:

cos(A-B) - cosC = 2cosBcosC

Step-by-Step 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