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Question: cos(A+B).cos(A-B) is identically equal to: (a) \({{\sin }^{2}}A-{{\sin }^{2}}B\) (b) \({{\cos }...

cos(A+B).cos(A-B) is identically equal to:
(a) sin⁡2A−sin⁡2B{{\sin }^{2}}A-{{\sin }^{2}}B
(b) cos⁡2A−sin⁡2B{{\cos }^{2}}A-{{\sin }^{2}}B
(c) cos⁡2A−cos⁡2B{{\cos }^{2}}A-{{\cos }^{2}}B
(d) cos2A.sin2B

Explanation

Solution

Hint: Start by applying the formula of cos(A+B) and cos(A-B) followed by factorisation at the end to reach the required expression.

Complete step-by-step answer:
Before starting with the solution to the above question, let us first discuss the important trigonometric formulas that we are going to be using in our solution.
cos⁡(X+Y)=cosX⁡cosY⁡−sinX⁡sin⁡Y.\cos \left( X+Y \right)=\operatorname{cosX}\operatorname{cosY}-\operatorname{sinX}\sin Y.
cos⁡(X−Y)=cosX⁡cosY⁡+sinX⁡sinY⁡.\cos \left( X-Y \right)=\operatorname{cosX}\operatorname{cosY}+\operatorname{sinX}\operatorname{sinY}.
cos⁡2X=1−sin⁡2X{{\cos }^{2}}X=1-{{\sin }^{2}}X .
sin⁡2X=1−cos⁡2X{{\sin }^{2}}X=1-{{\cos }^{2}}X .

We are now moving to the expression given in the question.
cos(A+B).cos(A-B).
When we apply the formula of cos(X+Y) and cos(X-Y) to the expression, we get
(cos⁡Acos⁡B−sin⁡Asin⁡B)(cos⁡Acos⁡B+sin⁡Asin⁡B)\left( \cos A\cos B-\sin A\sin B \right)\left( \cos A\cos B+\sin A\sin B \right)

Now moving on to multiplying and opening the brackets, the above expression becomes,
cos⁡2Acos⁡2B+cos⁡Acos⁡Bsin⁡Asin⁡B−cos⁡Acos⁡Bsin⁡Asin⁡B−sin⁡2Asin⁡2B{{\cos }^{2}}A{{\cos }^{2}}B+\cos A\cos B\sin A\sin B-\cos A\cos B\sin A\sin B-{{\sin }^{2}}A{{\sin }^{2}}B
=cos⁡2Acos⁡2B−sin⁡2Asin⁡2B={{\cos }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A{{\sin }^{2}}B

Now on applying the formula cos⁡2X=1−sin⁡2X{{\cos }^{2}}X=1-{{\sin }^{2}}X and sin⁡2X=1−cos⁡2X{{\sin }^{2}}X=1-{{\cos }^{2}}X, our expression becomes:
(1−sin⁡2A)cos⁡2B−sin⁡2A(1−cos⁡2B)\left( 1-{{\sin }^{2}}A \right){{\cos }^{2}}B-{{\sin }^{2}}A\left( 1-{{\cos }^{2}}B \right)
Again multiplying and opening the brackets, the expression comes out to be:
cos⁡2B−sin⁡2Acos⁡2B−sin⁡2A+cos⁡2Bsin⁡2A{{\cos }^{2}}B-{{\sin }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A+{{\cos }^{2}}B{{\sin }^{2}}A
=cos⁡2B−sin⁡2A={{\cos }^{2}}B-{{\sin }^{2}}A

But no option is matching with our final expression.
So, again using the formula cos⁡2X=1−sin⁡2X{{\cos }^{2}}X=1-{{\sin }^{2}}X and sin⁡2X=1−cos⁡2X{{\sin }^{2}}X=1-{{\cos }^{2}}X to convert the expression to a suitable form. On doing so our equation becomes:
1−sin⁡2B−(1−cos⁡2A)1-{{\sin }^{2}}B-\left( 1-{{\cos }^{2}}A \right)
=1−sin⁡2B−1+cos⁡2A=1-{{\sin }^{2}}B-1+{{\cos }^{2}}A
=1−sin⁡2B−1+cos⁡2A=1-{{\sin }^{2}}B-1+{{\cos }^{2}}A
=cos⁡2A−sin⁡2B={{\cos }^{2}}A-{{\sin }^{2}}B
So, we can say that cos(A+B).cos(A-B) is identically equal to cos⁡2A−sin⁡2B{{\cos }^{2}}A-{{\sin }^{2}}B .
Hence, option (b) is the correct answer.

Note: It is preferred to keep the expression as neat as possible by eliminating all the removable terms as the larger is the number of terms in your expression more is the probability of making a mistake. In trigonometry, it is often seen that many formulas lead to the same answer. However, the complexity of the question varies with the selection of the formula. The wiser the selection of formula, the easier the question will be to solve.