Question

If \[A+B+C=\pi \], and cos A = cos B . cos C, then tan B . tan C has the value equal to:
(a) 1
(b) \[\dfrac{1}{2}\]
(c) 2
(d) 3

Answer 1

Hint: To solve this question, we will find the value of sin A in terms of sines and cosines of B and C. After doing this, we will find the value of tan A with the help of the value of sin A obtained above. After this, we will find the value of tan A in terms of tan B and tan C and then we will equate both the values of tan A.

Complete step-by-step answer:
In this question, we have to find the value of tan B . tan C with the help of the data given in the question. In the question, it is given that
\[A+B+C=\pi \]
\[\Rightarrow B+C=\pi -A.....\left( i \right)\]
Now, we will write sin A in terms of sines and cosines of B and C. For this, we will find the value of sin (B + C) first, let its value be y. Thus,
\[y=\sin \left( B+C \right).....\left( ii \right)\]
Now, in the equation (ii) and (iii), we will use another identity which is as shown below.
\[\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y\]
After applying the above identity, we will get the following equation.
\[y=\sin B\cos C+\cos B\sin C\]
From (i) and (ii), we have the following equation.
\[y=\sin \left( \pi -A \right)\]
\[y=\sin A....\left( iv \right)\left[ \sin \left( {{180}^{o}}-\theta \right)=\sin \theta \right]\]
From (ii) and (iv), we will get the following equation,
\[\sin B\cos C+\sin C\cos B=\sin A.....\left( v \right)\]
Now, we have to find the value of tan A with the help of the above equation. So to obtain the value of tan A, we will divide the equation (v) with cos A. After doing this, we will get the following equation.
\[\dfrac{\sin B\cos C}{\cos A}+\dfrac{\sin C\cos B}{\cos A}=\dfrac{\sin A}{\cos A}\]
\[\dfrac{\sin B\cos C}{\cos A}+\dfrac{\sin C\cos B}{\cos A}=\tan A.....\left( vi \right)\]
Now, we know that cos A = cos B cos C (from the question). We will put the value of cos A in equation (vi). After doing this, we will get,
\[\dfrac{\sin B\cos C}{\cos B\cos C}+\dfrac{\sin C\cos B}{\cos B\cos C}=\tan A\]
\[\dfrac{\sin B}{\cos B}+\dfrac{\sin C}{\cos C}=\tan A\]
\[\tan B+\tan C=\tan A.....\left( vii \right)\]
Now, we will apply tan on both sides of the equation (i). After doing this, we will get the following equation.
\[\tan \left( B+C \right)=\tan \left( \pi -A \right)\]
\[\tan \left( B+C \right)=-\tan A\left[ \tan \left( \pi -\theta \right)=-\tan \theta \right]\]
\[\tan A=-\tan \left( B+C \right).....\left( viii \right)\]
From equation (vii) and (viii), we will get the following equation:
\[\tan B+\tan C=-\tan \left( B+C \right)\]
Now, we will apply another identity here:
\[\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\]
Thus, we will get the following equation:
\[\tan B+\tan C=-\left[ \dfrac{\tan B+\tan C}{1-\left( \tan B\tan C \right)} \right]\]
\[\dfrac{\tan B+\tan C}{\tan B+\tan C}=\dfrac{-1}{1-\tan B\tan C}\]
\[\Rightarrow 1=\dfrac{-1}{1-\tan B\tan C}\]
\[\Rightarrow 1-\tan B\tan C=-1\]
\[\Rightarrow \tan B.\tan C=1+1\]
\[\Rightarrow \tan B.\tan C=2\]
Hence, the option (c) is the right answer.

Note: The above question can also be solved as shown:
\[B+C=\pi -A\]
\[\cos \left( B+C \right)=\cos \left( \pi -A \right)\]
Here, we will apply the identity:
\[\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y\]
\[\cos \left( \pi -A \right)=-\cos A\]
\[\Rightarrow \cos B\cos C-\sin B\sin C=-\cos A\]
Dividing the whole equation by cos B cos C, we get,
\[1-\tan B\tan C=\dfrac{-\cos A}{\cos B\cos C}\]
\[1-\tan B\tan C=-1\]
\[\tan B\tan C=2\]