Question

If $\cos \left( A-B \right)=\dfrac{3}{5}$ , and $\tan A\tan B=2$ , then

$$\left( A \right)\cos A\cos B=-\dfrac{1}{5} \\\ \left( B \right)\sin A\sin B=-\dfrac{2}{5} \\\ \left( C \right)\cos \left( A+B \right)=-\dfrac{1}{5} \\\ \left( D \right)\sin A\cos B=\dfrac{4}{5} \\\$$

Answer 1

Hint : In order to solve the question, first of all, we need to look at the information given, $\cos \left( A-B \right)=\dfrac{3}{5}$ , this equation is simplified using the formula and the value of $\tan A\tan B=2$ is used in the other step , the values of $\sin A\sin B$ and $\cos A\cos B$ are found and from that $\cos \left( A+B \right)$ is also found.
Formula used: The formula that have been used in this question are, to simplify $\cos \left( A-B \right)=\dfrac{3}{5}$
$\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$
After finding the values of $\cos A\cos B$ and $\sin A\sin B$ , the values are put into the formula
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
And the value is found for $\cos \left( A+B \right)$
The formula for the tangent function is given as
$\tan A=\dfrac{\sin A}{\cos A}$

Complete step by step solution:
In order to solve the question, we simplify the equation given and use the relation $\tan A\tan B=2$ in other steps
Now, simplifying the equation $\cos \left( A-B \right)=\dfrac{3}{5}$ ,we use the formula for the cosine of difference of angles,
We get
$\Rightarrow \cos A\cos B+\sin A\sin B=\dfrac{3}{5}$
Dividing both the sides by $\cos A\cos B$

$$\Rightarrow \dfrac{\cos A\cos B}{\cos A\cos B}+\dfrac{\sin A\sin B}{\cos A\cos B}=\dfrac{3}{5\times \cos A\cos B} \\\ \Rightarrow 1+\tan A\tan B=\dfrac{3}{5\times \cos A\cos B} \\\$$

Now as it is given that $\tan A\tan B=2$
We get,

$$\Rightarrow 1+\tan A\tan B=\dfrac{3}{5\times \cos A\cos B} \\\ \Rightarrow 1+2=\dfrac{3}{5\times \cos A\cos B} \\\ \Rightarrow 3=\dfrac{3}{5\times \cos A\cos B} \\\ \Rightarrow \cos A\cos B=\dfrac{1}{5} \\\$$

Now dividing this equation
$\cos A\cos B + \sin A\sin B=\dfrac{3}{5}$ by $\sin A\sin B$

$$\Rightarrow \dfrac{\cos A\cos B}{\sin A\sin B}+\dfrac{\sin A\sin B}{\sin A\sin B}=\dfrac{3}{5\times \sin A\sin B} \\\ \Rightarrow \dfrac{1}{\tan A\tan B}+1=\dfrac{3}{5\times \sin A\sin B} \\\$$

Now, Putting $\tan A\tan B=2$

$$\Rightarrow \dfrac{1}{2}+1=\dfrac{3}{5\times \sin A\sin B} \\\ \Rightarrow \dfrac{3}{2}=\dfrac{3}{5\times \sin A\sin B} \\\ \Rightarrow \sin A\sin B=\dfrac{2}{5} \\\$$

Now as we know, $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
Therefore, $\cos \left( A+B \right)=\dfrac{1}{5}-\dfrac{2}{5}=-\dfrac{1}{5}$
Thus, the answer is $\cos \left( A+B \right)=-\dfrac{1}{5}$ , i.e. option $\left( C \right)$ is correct, all the other options are not correct , because we are getting the values other than that given in the options.
So, the correct answer is “Option C”.

Note : It is very important to check the correctness of all the options by finding the values for the relations given in options, getting an answer other than the option as given may confuse you, but solve more to get the validation of all the options and solve till we get the value for the relation $\cos \left( A+B \right)$ .