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

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:
$\begin{aligned} & \sin \left( A+B \right)=\sin \left( {{180}^{\circ }}-C \right) \\\ & \sin \left( A+B \right)=\sin C \\\ \end{aligned}$
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:
$\begin{aligned} & \left( \cot A+\cot B \right)=\left( \dfrac{\cos A}{\sin A}+\dfrac{\cos B}{\sin B} \right) \\\ & =\dfrac{\cos A\sin B+\cos B\sin A}{\sin A\sin B}......(1) \end{aligned}$
Similarly, for $\left( \cot B+\cot C \right)$
$\begin{aligned} & \left( \cot B+\cot C \right)=\left( \dfrac{\cos B}{\sin B}+\dfrac{\cos C}{\sin C} \right) \\\ & =\dfrac{\cos B\sin C+\cos C\sin B}{\sin B\sin C}......(2) \end{aligned}$
And for $\left( \cot A+\cot C \right)$
$\begin{aligned} & \left( \cot A+\cot C \right)=\left( \dfrac{\cos A}{\sin A}+\dfrac{\cos C}{\sin C} \right) \\\ & =\dfrac{\cos A\sin C+\cos C\sin A}{\sin A\sin C}......(3) \end{aligned}$

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:
$\begin{aligned} & \sin \left( A+B \right)=\sin \left( {{180}^{\circ }}-C \right) \\\ & \sin \left( A+B \right)=\sin C \\\ \end{aligned}$
$\left[ \because \sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta \right]$

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,

& \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{aligned}$$ Now, multiply equation (7), (8) and (9), we get: $$$$$$\begin{aligned} & \left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right) \\\ & =\dfrac{\sin A}{\sin B\sin C}\times \dfrac{\sin B}{\sin A\sin C}\times \dfrac{\sin C}{\sin A\sin B} \\\ & =\dfrac{1}{\sin A\sin B\sin C}......(10) \\\ \end{aligned}$$ Since, $$\dfrac{1}{\sin \theta }=\text{cosec }\theta $$ So, we can write equation (10) as: $$\begin{aligned} & \left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right) \\\ & =\cos \text{ec}A\cos \text{ec}B\cos \text{ec}C \\\ \end{aligned}$$ **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: $$$$$$\begin{aligned} & \left( \cot \text{B}+\cot \text{C} \right)\left( \cot \text{C}+\cot \text{A} \right)\left( \cot \text{A}+\cot \text{B} \right) \\\ & =\left( \cot {{60}^{\circ }}+\cot {{60}^{\circ }} \right)\left( \cot {{60}^{\circ }}+\cot {{60}^{\circ }} \right)\left( \cot {{60}^{\circ }}+\cot {{60}^{\circ }} \right) \\\ & =\left( \dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}} \right)\left( \dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}} \right)\left( \dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}} \right) \\\ & =\dfrac{\left( 2\sqrt{3} \right)\left( 2\sqrt{3} \right)\left( 2\sqrt{3} \right)}{3\sqrt{3}} \\\ & =8 \\\ \end{aligned}$$ Now, go for the given options. Check if any option satisfies with the given angle.