Question

If $A+B+C={{180}^{\circ }}$, then the value of \[\left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right)\]will be
a) $\sec \text{A}\sec \text{B}\sec \text{C}$
b) $\text{cosecA cosecB cosecC}$
c) $\tan \text{A}\tan \text{B}\tan \text{C}$
c) 1

Answer 1

Hint: Since we are given that: $A+B+C={{180}^{\circ }}$. We can write, $A+B={{180}^{\circ }}-C$. Apply sine on both sides of the equation, we get:
$\begin{align}
  & \sin \left( A+B \right)=\sin \left( {{180}^{\circ }}-C \right) \\
 & \sin \left( A+B \right)=\sin C \\
\end{align}$
Since we need to find the values of cotangent of given angles, we can find value of cosine and sine of given angles and divide them.
Therefore, we can write:
$\cot A=\dfrac{\cos A}{\sin A};\cot B=\dfrac{\cos B}{\sin B};\cot C=\dfrac{\cos C}{\sin C}$
Substitute the values in the expression \[\left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right)\] and solve the equation to get the value.

Complete step by step answer:
We have the following expression to solve: \[\left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right)\]
Let us solve for $\left( \cot A+\cot B \right)$first.
We can write this expression as:
$\begin{align}
  & \left( \cot A+\cot B \right)=\left( \dfrac{\cos A}{\sin A}+\dfrac{\cos B}{\sin B} \right) \\
 & =\dfrac{\cos A\sin B+\cos B\sin A}{\sin A\sin B}......(1)
\end{align}$
Similarly, for $\left( \cot B+\cot C \right)$
$\begin{align}
  & \left( \cot B+\cot C \right)=\left( \dfrac{\cos B}{\sin B}+\dfrac{\cos C}{\sin C} \right) \\
 & =\dfrac{\cos B\sin C+\cos C\sin B}{\sin B\sin C}......(2)
\end{align}$
And for $\left( \cot A+\cot C \right)$
$\begin{align}
  & \left( \cot A+\cot C \right)=\left( \dfrac{\cos A}{\sin A}+\dfrac{\cos C}{\sin C} \right) \\
 & =\dfrac{\cos A\sin C+\cos C\sin A}{\sin A\sin C}......(3)
\end{align}$

By applying identity: \[\left( \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B \right)\] in equation (1), we can write:
\[\left( \cot A+\cot B \right)=\dfrac{\sin \left( A+B \right)}{\sin A\sin B}......(4)\]
Similarly, applying the sine-addition identity in equation (2) and (3), we get:
\[\left( \cot B+\cot C \right)=\dfrac{\sin \left( B+C \right)}{\sin B\sin C}......(5)\]
\[\left( \cot A+\cot C \right)=\dfrac{\sin \left( A+C \right)}{\sin A\sin C}......(6)\]

Since we are given that: $A+B+C={{180}^{\circ }}$. We can write, $A+B={{180}^{\circ }}-C$.
Apply sine on both sides of the equation, we get:
$\begin{align}
  & \sin \left( A+B \right)=\sin \left( {{180}^{\circ }}-C \right) \\
 & \sin \left( A+B \right)=\sin C \\
\end{align}$
$\left[ \because \sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta \right]$

Substitute the value of $\sin \left( A+B \right)$ in equation (4), we get:
\[\left( \cot A+\cot B \right)=\dfrac{\sin C}{\sin A\sin B}......(7)\]
Similarly,
\[\begin{align}
  & \left( \cot B+\cot C \right)=\dfrac{\sin A}{\sin B\sin C}......(8) \\
 & \left( \cot A+\cot C \right)=\dfrac{\sin B}{\sin A\sin C}......(9) \\
\end{align}\]

Now, multiply equation (7), (8) and (9), we get:
\[\]\[\begin{align}
  & \left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right) \\
 & =\dfrac{\sin A}{\sin B\sin C}\times \dfrac{\sin B}{\sin A\sin C}\times \dfrac{\sin C}{\sin A\sin B} \\
 & =\dfrac{1}{\sin A\sin B\sin C}......(10) \\
\end{align}\]

Since, \[\dfrac{1}{\sin \theta }=\text{cosec }\theta \]
So, we can write equation (10) as:
\[\begin{align}
  & \left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right) \\
 & =\cos \text{ec}A\cos \text{ec}B\cos \text{ec}C \\
\end{align}\]

So, the correct answer is “Option B”.

Note: There is another method to solve the expression. We can assume that A = B = C. So, we get the value of A, B and C = \[{{60}^{\circ }}\]. Put the values of cotangent of \[{{60}^{\circ }}\]and solve the expression. For the given expression: \[\left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right)\], for \[\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}}\], we can write:
\[\]\[\begin{align}
  & \left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right) \\
 & =\left( \cot {{60}^{\circ }}+\cot {{60}^{\circ }} \right)\left( \cot {{60}^{\circ }}+\cot {{60}^{\circ }} \right)\left( \cot {{60}^{\circ }}+\cot {{60}^{\circ }} \right) \\
 & =\left( \dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}} \right)\left( \dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}} \right)\left( \dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}} \right) \\
 & =\dfrac{\left( 2\sqrt{3} \right)\left( 2\sqrt{3} \right)\left( 2\sqrt{3} \right)}{3\sqrt{3}} \\
 & =8 \\
\end{align}\]
Now, go for the given options. Check if any option satisfies with the given angle.