Question

If $A + B + C = \pi$ and $\cos A = \cos B\cos C$, then $\tan B\tan C$ is equal to
A) $\dfrac{1}{2}$
B) $2$
C) $1$
D) $- \dfrac{1}{2}$

Answer 1

Here, we will first find the value of $A$ in terms of $B,C$. Then we will put the value of $A$ in the equation $\cos A = \cos B\cos C$. Then we will use the trigonometry formula to expand the equation. Then we will solve the equation to get the value of $\tan B\tan C$.

Complete step by step solution:
It is given that $A + B + C = \pi$ and $\cos A = \cos B\cos C$.
First, we will take the equation $\cos A = \cos B\cos C$ and put the value of $A$ in terms of the $B,C$ with the help of the equation $A + B + C = \pi$. Therefore, we get
$\Rightarrow A = \pi - \left( {B + C} \right)$
Now substituting the value of $A$ in the equation $\cos A = \cos B\cos C$, we get
$\Rightarrow \cos \left( {\pi - \left( {B + C} \right)} \right) = \cos B\cos C$
We can see that the function $\cos \left( {\pi - \left( {B + C} \right)} \right)$ is in the second quadrant and we know that the value of the cos function is negative in the second quadrant i.e. $\cos \left( {\pi - \theta } \right) = - \cos \theta$. Therefore, we get
$\Rightarrow - \cos \left( {B + C} \right) = \cos B\cos C$
Now we will use the formula $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$. Therefore, we get
$\Rightarrow - \left( {\cos B\cos C - \sin B\sin C} \right) = \cos B\cos C$
Now we will solve this equation to get the value of $\tan B\tan C$. Therefore, we get
$\Rightarrow - \cos B\cos C + \sin B\sin C = \cos B\cos C$
$\Rightarrow \sin B\sin C = \cos B\cos C + \cos B\cos C = 2\cos B\cos C$
Now we will take the cos terms on the other side of the equation, we get
$\Rightarrow \dfrac{{\sin B\sin C}}{{\cos B\cos C}} = 2$
We know that the ratio of the sine to the cos is equals to the tan function i.e. $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$, we get
$\Rightarrow \tan B\tan C = 2$
Hence, $\tan B\tan C$ is equal to 2.

So, option B is the correct option.

Note:
We should know the different properties of the trigonometric function. In the first quadrant all the functions i.e. sin, cos, tan, cot, sec, cosec is positive. In the second quadrant, only the sin and cosec function are positive and all the other functions are negative. In the third quadrant, only tan and cot function is positive and in the fourth quadrant, only cos and sec function is positive. Also, we should know the basic properties of the trigonometric functions and basic trigonometric formulas.
$\begin{array}{l}\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\\\\\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\\\\\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\\\ \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B\end{array}$