Question

# Prove that: $\dfrac{a\sec \text{A + }b\sec \text{B + }c\sec \text{C}}{\operatorname{tanA}\cdot tanB\cdot tanC}\text{ = 2R}$

Hint: As A, B and C are the angles of a triangle, use the property the sum of the three angles of a triangle is equal to${{180}^{\circ }}$. Now adjust any one of the angles on the right hand side and take tan on both sides of the equation. Then use the law of sines $\dfrac{a}{\sin \text{A}}\text{ = }\dfrac{b}{\operatorname{sinB}}\text{ = }\dfrac{c}{\operatorname{sinC}}\text{ = 2R}$, where a, b and c are the three sides of the triangle and R is the circumradius of the triangle, to prove $\dfrac{a\sec \text{A + }b\sec \text{B + }c\sec \text{C}}{\operatorname{tanA}\cdot tanB\cdot tanC}\text{ = 2R}$.

We know that A, B and C are the angles of a triangle. Thus, we can write
$\text{A + B + C = 18}{{\text{0}}^{\circ }}$
Subtracting the angle from both sides of the equation, we get,
$\text{A + B = 18}{{\text{0}}^{\circ }}-\text{ C}$
Taking tan on both sides of the equation, we get,
\begin{align} & \tan \left( \text{A+B} \right)\text{ = tan}\left( {{180}^{\circ }}-\text{C} \right) \\ & \Rightarrow \text{ }\dfrac{\tan \text{A + tanB}}{1-\tan \text{A}\tan \text{B}}\text{ = }-\tan \text{C} \\ & \Rightarrow \text{ tanA + tanB = }-\text{tanC}\left( 1-\tan \text{A tanB} \right) \\ & \Rightarrow \text{ tanA + tanB = }-\tan \text{C + tanA tanB tanC} \\ & \Rightarrow \text{ tanA + tanB + tanC = tanA tanB tanC} \\ & \therefore \text{ }\dfrac{\text{sinA}}{\cos \text{A}}\text{ + }\dfrac{\text{sinB}}{\operatorname{cosB}}\text{ + }\dfrac{\text{sinC}}{\operatorname{cosC}}\text{ = tanA tanB tanC }....\left( \text{i} \right) \\ \end{align}
Now, we know the law of sines, which states that,
$\dfrac{a}{\sin \text{A}}\text{ = }\dfrac{b}{\operatorname{sinB}}\text{ = }\dfrac{c}{\operatorname{sinC}}\text{ = 2R}$
Therefore, the sine of angles A, B and C can be expressed in the form
\begin{align} & \sin \text{A = }\dfrac{\text{a}}{\text{2R}}\text{ }....\text{(ii)} \\ & \operatorname{sinB}\text{ = }\dfrac{\text{b}}{\text{2R}}\text{ }....\text{(iii)} \\ & \operatorname{sinC}\text{ = }\dfrac{\text{c}}{\text{2R}}\text{ }....\text{(iv)} \\ \end{align}
Thus, putting the values of the sine of the angles form equations (ii), (iii) and (iv) in the equation (i), we get,
\begin{align} & \dfrac{\text{sinA}}{\cos \text{A}}\text{ + }\dfrac{\text{sinB}}{\operatorname{cosB}}\text{ + }\dfrac{\text{sinC}}{\operatorname{cosC}}\text{ = tanA tanB tanC} \\ & \Rightarrow \text{ }\dfrac{\text{a}}{\text{2RcosA}}\text{ + }\dfrac{\text{b}}{\text{2R cosB}}\text{ + }\dfrac{\text{c}}{\text{2R cosC}}\text{ = tanA tanB tanC} \\ & \Rightarrow \text{ }\dfrac{\text{a secA + b secB + c secC}}{\text{2R}}\text{ = tanA tanB tanC} \\ & \therefore \text{ }\dfrac{\text{a secA + b secB + c secC}}{\text{tanA tanB tanC}}\text{ = 2R } \\ \end{align}
Hence, it is proved that $\dfrac{a\sec \text{A + }b\sec \text{B + }c\sec \text{C}}{\operatorname{tanA}\cdot tanB\cdot tanC}\text{ = 2R}$.

Note: In any triangle ABC, it is a known property that tanA + tanB + tanC = tanA tanB tanC. So one can also use that directly, instead of proving the property, to prove the statement given in the question. Also, tan($\text{18}{{\text{0}}^{\circ }}-\text{ C}$) is negative because the angle belongs to the second quadrant and the value of tan of any angle in the second quadrant is always negative.