Question

In ∆ABC, prove that:

$\frac{b\sec B+c\sec C}{\tan B+\tan C}=\frac{c\sec C+a\sec A}{\tan C+\tan A}=\frac{a\sec A+b\sec B}{\tan A+\tan B}$

Answer 1

Let ABC be any triangle.

Suppose $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Now,
$$\begin{gathered} \frac{b\sec B+c\sec C}{\tan B+\tan C}=\frac{{\displaystyle \frac{b}{\cos B}}+{\displaystyle \frac{c}{\cos C}}}{{\displaystyle \frac{\sin B}{\cos B}}+{\displaystyle \frac{\sin C}{\cos C}}} \\ =\frac{b\cos C+c\cos B}{\sin B\cos C+\sin C\cos B} \\ =\frac{k\sin B\cos C+k\sin C\cos B}{\sin B\cos C+\sin C\cos B} \\ =\frac{k\left(\sin B\cos C+\sin C\cos B\right)}{\sin B\cos C+\sin C\cos B}=k...\left(1\right) \end{gathered}$$
Also,
$$\begin{gathered} \frac{c\sec C+a\sec A}{\tan C+\tan A}=\frac{{\displaystyle \frac{c}{\cos C}}+{\displaystyle \frac{a}{\cos A}}}{{\displaystyle \frac{\sin C}{\cos C}}+{\displaystyle \frac{\sin A}{\cos A}}} \\ =\frac{c\cos A+a\cos C}{\sin C\cos A+\sin A\cos C} \\ =\frac{k\sin C\cos A+k\sin A\cos C}{\sin C\cos A+\sin A\cos C} \\ =\frac{k\left(\sin C\cos A+\sin A\cos C\right)}{\sin C\cos A+\sin A\cos C}=k...\left(2\right) \\ \mathrm{and} \\ \frac{a\sec A+b\sec B}{\tan A+\tan B}=\frac{{\displaystyle \frac{a}{\cos A}}+{\displaystyle \frac{b}{\cos B}}}{{\displaystyle \frac{\sin A}{\cos A}}+{\displaystyle \frac{\sin B}{\cos B}}} \\ =\frac{a\cos B+b\cos A}{\sin A\cos B+\sin B\cos A} \\ =\frac{k\left(\sin A\cos B+\sin B\cos A\right)}{\sin A\cos B+\sin B\cos A}=k...\left(3\right) \end{gathered}$$

From (1), (2) and (3), we get:

$\frac{b\sec B+c\sec C}{\tan B+\tan C}=\frac{c\sec C+a\sec A}{\tan C+\tan A}=\frac{a\sec A+b\sec B}{\tan A+\tan B}$

Hence proved.