Question

Q. cos(A+B)+cosC=0 Show that?

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

cos(A+B)+cosC=0

Step-by-step explanation:

Consider the triangle ABC.

$\angle A+\angle B+\angle C=180^0$

In simple, we can write above as,

A+B+C = 180

i.e., A+B = 180-C

Taking Cos on both sides, we get

cos(A+B) =  cos(180-C)------------( 1 )

Now consider LHS(Left hand side) from the question

cos(A+B)+cos C

                  = cos (180-C) +cos (C)

                  = -cos (C) + cos(C)

                  = 0

                  = RHS (Right Hand side)

Hence Proved.