Question

If \[\cos (A - B) = \dfrac{3}{5}\] and \[\tan A\tan B = 2\] , then choose the correct option.
     A. \[\cos A\cos B = \dfrac{1}{5}\]
    B.\[\sin A\sin B = - \dfrac{2}{5}\]
    C.\[\cos (A + B) = - \dfrac{1}{5}\]
    D.\[\sin A\cos B = \dfrac{4}{5}\]

Answer 1

Hint: Write \[\tan A\tan B\] in terms of \[\sin A\],\[\sin B\],\[\cos A\] and \[\cos B\].From there we will get the relation in \[\sin A\sin B\] and \[\cos A\cos B\]. Expand the formulae of \[\cos (A - B)\] and use the relations obtained earlier to get the desired values.

Formula Used
1.\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
2.\[\cos (A - B) = \cos A\cos B + \sin A\sin B\]
3. \[\cos (A + B) = \cos A\cos B - \sin A\sin B\]

Complete step by step solution
Given- \[\tan A\tan B = 2\]
We know that \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] ,using this identity in above relation
 \[\dfrac{{\sin A\sin B}}{{\cos A\cos B}} = 2\]
Cross-multiplying the terms
\[\sin A\sin B = 2\cos A\cos B\] (1)
Again, \[\cos (A - B) = \dfrac{3}{5}\] (Given)
Applying the formula \[\cos (A - B) = \cos A\cos B + \sin A\sin B\]
\[\cos A\cos B + \sin A\sin B = \dfrac{3}{5}\]
Using equation (1)
\[\cos A\cos B + 2\cos A\cos B = \dfrac{3}{5}\]
\[ \Rightarrow 3\cos A\cos B = \dfrac{3}{5}\]
\[ \Rightarrow \cos A\cos B = \dfrac{1}{5}\]
Again, using equation (1)
\[\sin A\sin B = 2\cos A\cos B\]
\[ \Rightarrow \sin A\sin B = 2*\dfrac{1}{5}\]
\[ \Rightarrow \sin A\sin B = \dfrac{2}{5}\]
Now using the formula \[\cos (A + B) = \cos A\cos B - \sin A\sin B\]
\[\cos (A + B) = \dfrac{1}{5} - \dfrac{2}{5}\]
\[ \Rightarrow \cos (A + B) = - \dfrac{1}{5}\]
Hence option C is the correct answer.

Note: The values \[\sin A\cos B\] and \[\cos A\sin B\] are not given. Thus we will give equations to find the values of \[\sin A\cos B\] and \[\cos A\sin B\].
Students must not get confused in the sum and difference formulas of cosine; they are as follows –
\[\cos (A - B) = \cos A\cos B + \sin A\sin B\]
\[\cos (A + B) = \cos A\cos B - \sin A\sin B\]