Question

In a $\Delta ABC$ ,
Prove that:-
$\cos (A+B)+\cos (C)=0$

Answer 1

Hint: -In such question, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.

Complete step-by-step answer:
The below mentioned formula may be used in the solution which is as follows
In $\Delta ABC$ ,
$A+B+C={{180}^{\circ }}$
(As sum of the angles of a triangle is equal to ${{180}^{\circ }}$ )
Another important result that will be used in solving the question is as follows
$\cos ({{180}^{\circ }}-\theta )=-\cos (\theta )$
Now, these are the results that would be used to prove the proof mentioned in this question as using these identities, we would convert the left hand side that is L.H.S. or the right hand side that is R.H.S. to make either of them equal to the other.
In this particular question, we will first modify the sum of A and B as ${{180}^{\circ }}-C$ using the above fact. Then we can simplify further to get the answer.
As mentioned in the question, we have to prove the given expression.
Now, we will start with the left hand side that is L.H.S. and try to make the necessary changes that are given in the hint, first, as follows

& =\cos (A+B)+\cos (C) \\\ & =\cos ({{180}^{\circ }}-C)+\cos (C) \\\ \end{aligned}$$ Now, on simplifying the angle of the trigonometric function cos, we get the following result $$=-\cos C+\cos C$$ (Using the identities that are mentioned in the hint) Now, on cancelling the terms, we get the following $$=0$$ Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved. Note: -Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof.