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

Answer

$\frac{\text{b}\sec\text{B + c}\sec\text{C}}{\tan\text{B}+\tan\text{C}}=\frac{\text{c}\sec\text{C + a}\sec\text{A}}{\tan\text{C}+\tan\text{A}}=\frac{\text{a}\sec\text{A + b}\sec\text{B}}{\tan\text{A}+\tan\text{B}}$
$\frac{\text{b}\sec\text{B + c}\sec\text{C}}{\tan\text{B}+\tan\text{C}}$
$=\frac{\text{k}\sin\text{B}\sec\text{B + k}\sin\text{C}\sec\text{C}}{\tan\text{B}+\tan\text{C}}$
$=\frac{\text{k}\sin\text{B}\frac{1}{\cos\text{B}}+\text{k}\sin\text{C}\frac{1}{\cos\text{C}}}{\tan\text{B}+\tan\text{C}}$
$=\frac{\text{k}\tan\text{B + k}\tan\text{C}}{\tan\text{B}+\tan\text{C}}=\frac{\text{k}(\tan\text{B}+\tan\text{C})}{\tan\text{B}+\tan\text{C}}=\text{k}$
Similarly, $\frac{\text{c}\sec\text{C}+\text{a}+\sec\text{A}}{\tan\text{C}+\tan\text{A}}=\text{k}$
Similarly, $\frac{\text{a}\sec\text{A}+\text{b}+\sec\text{B}}{\tan\text{A}+\tan\text{B}}=\text{k}$
Hence, it is proved that
$\frac{\text{b}\sec\text{B + c}\sec\text{C}}{\tan\text{B}+\tan\text{C}}=\frac{\text{c}\sec\text{C + a}\sec\text{A}}{\tan\text{C}+\tan\text{A}}=\frac{\text{a}\sec\text{A + b}\sec\text{B}}{\tan\text{A}+\tan\text{B}}=\text{k}$

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