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What is the value of $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$?
A . $\cos A$
B . 0
C . $\sqrt{3}\sin A$
D . $\sqrt{3}\cos A$

Answer
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Hint: In this question, we are given an identity $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$. First, we use the formula of $\cos (A+B)+\cos (A-B)$ to solve the question and then we use the complimentary angles of trigonometric ratios and simplify them to get the desirable answer.

Formula Used:
In this question, we use the identity:-
$\cos (A+B)+\cos (A-B)=2\cos A\cos B$
And the complimentary angle
$\cos ({{270}^{\circ }}-A)=-\sin A$ to solve the question.

Complete step- by- step Solution:
We have given an identity $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$………………………… (1)
Now we know the identity
$\cos (A+B)+\cos (A-B)=2\cos A\cos B$
$\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)=2\cos {{240}^{\circ }}\cos A$
Now by putting the value in equation (1), we get
$\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$ = $\cos A+2\cos {{240}^{\circ }}\cos A$
$\cos A+2\cos {{240}^{\circ }}\cos A$ = $\cos A(1+2\cos ({{270}^{\circ }}-{{30}^{\circ }}))$
We know the complimentary angles of trigonometric ratios the value of
$\cos ({{270}^{\circ }}-A)=-\sin A$
So $\cos ({{270}^{\circ }}-{{30}^{\circ }})=-\sin {{30}^{\circ }}$
Then $\cos A+2\cos {{240}^{\circ }}\cos A$ = $\cos A(1+2(-\sin {{30}^{\circ }}))$
And value of $\sin {{30}^{\circ }}=\dfrac{1}{2}$
Then the value of $\cos A+2\cos {{240}^{\circ }}\cos A$ = $\cos A(1-2\times \dfrac{1}{2}))$
Which is equal to $\cos A(1-1)$ = 0
Hence, $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$ = 0

Thus, Option ( B ) is the correct answer.

Note: In trigonometry, different type of questions can be solved with the help of different trigonometry formulas. These questions may include trigonometric ratios, Pythagoras theorem, product identities etc. There are many trigonometry formulas. We choose the formula according to our question demands. Learning and remembering these formulas help the student to solve the questions accurately and able to choose the correct option.