Question

In any triangle $ABC$, prove that:
$a\left( \cos B.\cos C+\cos A \right)=b\left( \cos C.\cos A+\cos B \right)=c\left( \cos A.\cos B+\cos C \right)$

Answer 1

Hint: We will take the part $a\left( \cos B.\cos C+\cos A \right)$ from the given condition. Here we will substitute the value of $a$ from sine rule $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=K$. Now we will multiply and divide the equation with two and apply the known formula$2\cos A.\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$. After that we will use the formula $A+B+C=\pi $ and simplify the obtained equation to get a constant value.

Complete step-by-step solution

Given that,
$a\left( \cos B.\cos C+\cos A \right)=b\left( \cos C.\cos A+\cos B \right)=c\left( \cos A.\cos B+\cos C \right)$
Taking the part $a\left( \cos B.\cos C+\cos A \right)$ separately.
From the sine rule $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=K$, the value of $a$ is $a=K\sin A$. Substituting the value of $a$ in the above equation, then we will get
$a\left( \cos B.\cos C+\cos A \right)=K\sin A\left( \cos B.\cos C+\cos A \right)$
Multiplying and dividing with $2$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K\sin A}{2}\times 2\left( \cos B.\cos C+\cos A \right) \\
 & \Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K\sin A}{2}\left( 2\cos B.\cos C+2\cos A \right) \\
\end{align}$
We know that $2\cos A.\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$, then
$\Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K\sin A}{2}\left( \cos \left( B+C \right)+\cos \left( B-C \right)+2\cos A \right)$
We know that the sum of the angles in a triangle is $180{}^\circ $. $\Rightarrow A+B+C=180{}^\circ \Rightarrow B+C=180{}^\circ -A$
Substituting the value of $B+C$ in the value of $a\left( \cos B.\cos C+\cos A \right)$ then we will get
$\Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K\sin A}{2}\left( \cos \left( 180{}^\circ -A \right)+\cos \left( B-C \right)+2\cos A \right)$
We know that $\cos \left( 180{}^\circ -\theta \right)=-\cos \theta $, then the value of $a\left( \cos B.\cos C+\cos A \right)$ is
$\begin{align}
  & \Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K\sin A}{2}\left( -\cos A+\cos \left( B-C \right)+2\cos A \right) \\
 & \Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K\sin A}{2}\left( \cos A+\cos \left( B-C \right) \right) \\
 & \Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K}{2}\left( \sin A.\cos A+\sin A.\cos \left( B-C \right) \right) \\
\end{align}$
Again, multiplying and dividing with $2$, then we will get
$\Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K}{2}\times \dfrac{1}{2}\left( 2\sin A.\cos A+2\sin A.\cos \left( B-C \right) \right)$
We have values $2\sin A.\cos A=\sin 2A$ and $\sin A=\sin \left( 180{}^\circ -\left( B+C \right) \right)=\sin \left( B+C \right)$, substituting these values in the above equation, then we will get
$\Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K}{4}\left( \sin 2A+2\sin \left( B+C \right).\cos \left( B-C \right) \right)$
We know that $2\sin \left( B+C \right).\cos \left( B-C \right)=\sin 2B+\sin 2C$, then
$\Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{K}{4}\left( \sin 2A+\sin 2B+\sin 2C \right)..\left( \text{i} \right)$
Similarly, we will get
$b\left( \cos C.\cos A+\cos B \right)=\dfrac{K}{4}\left( \sin 2A+\sin 2B+\sin 2C \right)$
And $c\left( \cos A.\cos B+\cos C \right)=\dfrac{K}{4}\left( \sin 2A+\sin 2B+\sin 2C \right)$
$\therefore a\left( \cos B.\cos C+\cos A \right)=b\left( \cos C.\cos A+\cos B \right)=c\left( \cos A.\cos B+\cos C \right)$
Hence proved.

Note: We can also solve the above problem in another method. We have the sum of the angles in a triangle as $A+B+C=180{}^\circ $. Now the value of $\cos \left( B+C \right)$ is
$\cos \left( B+C \right)=\cos \left( 180{}^\circ -A \right)$
Applying the formula $\cos \left( A+B \right)=\cos A.\cos B-\sin A.\sin B$ in the above equation, then we will get
$\begin{align}
  & \cos B.\cos C-\sin B.\sin C=-\cos A \\
 & \Rightarrow \cos B.\cos C+\cos A=\sin B.\sin C \\
\end{align}$
From the sine rule $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=4R$ the value of $\sin B.\sin C=\dfrac{bc}{{{\left( 4R \right)}^{2}}}$
Now the value of $a\left( \cos B.\cos C+\cos A \right)$ is given by
$\begin{align}
  & a\left( \cos B.\cos C+\cos A \right)=a\sin B.\sin C \\
 & \Rightarrow a\left( \cos B.\cos C+\cos A \right)=\dfrac{abc}{{{\left( 4R \right)}^{2}}} \\
\end{align}$
Similarly, we will have the values of $b\left( \cos C.\cos A+\cos B \right)=\dfrac{abc}{{{\left( 4R \right)}^{2}}}$ and $c\left( \cos A.\cos B+\cos C \right)=\dfrac{abc}{{{\left( 4R \right)}^{2}}}$
$\therefore a\left( \cos B.\cos C+\cos A \right)=b\left( \cos C.\cos A+\cos B \right)=c\left( \cos A.\cos B+\cos C \right)$
From both the methods we got the same result.