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Question
In ∆ABC, prove that: \[\frac{b \sec B + c \sec C}{\tan B + \tan C} = \frac{c \sec C + a \sec A}{\tan C + \tan A} = \frac{a \sec A + b \sec B}{\tan A + \tan B}\]
Solution
Let ABC be any triangle
Suppose
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]
Now,
\[ = \frac{b\cos C + c\cos B}{\sin B\cos C + \sin C\cos B}\]
\[ = \frac{k\sin B\cos C + k\sin C\cos B}{\sin B\cos C + \sin C\cos B} \]
\[ = \frac{k\left( \sin B\cos C + \sin C\cos B \right)}{\sin B\cos C + \sin C\cos B} = k . . . \left( 1 \right)\]
\[ = \frac{c\cos A + a\cos C}{\sin C\cos A + \sin A\cos C}\]
\[ = \frac{k\sin C\cos A + k\sin A\cos C}{\sin C\cos A + \sin A\cos C} \]
\[ = \frac{k\left( \sin C\cos A + \sin A\cos C \right)}{\sin C\cos A + \sin A\cos C} = k . . . \left( 2 \right)\]
\[\text{ and }\]
\[\frac{a\sec A + b\sec B}{\tan A + \tan B} = \frac{\frac{a}{\cos A} + \frac{b}{\cos B}}{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}\]
\[ = \frac{a\cos B + b\cos A}{\sin A\cos B + \sin B\cos A}\]
\[ = \frac{k\left( \sin A\cos B + \sin B\cos A \right)}{\sin A\cos B + \sin B\cos A} = k . . . \left( 3 \right)\]
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