In ∆ABC, prove that:

Let ABC be any triangle.

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

Now,

$\frac{b \sec B+c \sec C}{\tan B+\tan C}=\frac{\frac{b}{\cos B}+\frac{c}{\cos C}}{\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}}$

$=\frac{b \cos C+c \cos B}{\sin B \cos C+\sin C \cos B}$

$=\frac{k \sin B \cos C+k \sin C \cos B}{\sin B \cos C+\sin C \cos B}$

$=\frac{k(\sin B \cos C+\sin C \cos B)}{\sin B \cos C+\sin C \cos B}=k \quad \ldots(1)$

Also,

$\frac{c \sec C+a \sec A}{\tan C+\tan A}=\frac{\frac{c}{\cos C}+\frac{a}{\cos A}}{\frac{\sin C}{\cos C}+\frac{\sin A}{\cos A}}$

$=\frac{c \cos A+a \cos C}{\sin C \cos A+\sin A \cos C}$

$=\frac{k \sin C \cos A+k \sin A \cos C}{\sin C \cos A+\sin A \cos C}$

$=\frac{k(\sin C \cos A+\sin A \cos C)}{\sin C \cos A+\sin A \cos C}=k \quad \ldots(2)$

and,

$\frac{a \sec A+b \sec B}{\tan A+\tan B}=\frac{\frac{a}{\cos A}+\frac{b}{\cos B}}{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}$

$=\frac{a \cos B+b \cos A}{\sin A \cos B+\sin B \cos A}$

$=\frac{k(\sin A \cos B+\sin B \cos A)}{\sin A \cos B+\sin B \cos A}=k$      ....(3)

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.