Answer: C
\( \sin 51^\circ = \cfrac{a}{\sqrt{a^2 + b^2}} \) \(\therefore \csc 51^\circ = \cfrac{\sqrt{a^2 + b^2}}{a} \) i.e., \( \csc^2 51^\circ = \cfrac{a^2 + b^2}{a^2} \) i.e., \( \csc^2 51^\circ - 1 = \cfrac{a^2 + b^2}{a^2} - 1 \) i.e., \( \cot^2 51^\circ = \cfrac{\cancel{a^2} + b^2 - \cancel{a^2}}{a^2} \) i.e., \( \cot 51^\circ = \cfrac{b}{a} \) i.e., \( \tan 51^\circ = \cfrac{a}{b} \) \(\therefore \tan 51^\circ + \tan 39^\circ \) \(= \tan 51^\circ + \tan(90^\circ - 51^\circ) \) \(= \tan 51^\circ + \cot 51^\circ \) \(= \cfrac{a}{b} + \cfrac{b}{a} \) \(= \cfrac{a^2 + b^2}{ab} \)
\( \sin 51^\circ = \cfrac{a}{\sqrt{a^2 + b^2}} \) \(\therefore \csc 51^\circ = \cfrac{\sqrt{a^2 + b^2}}{a} \) i.e., \( \csc^2 51^\circ = \cfrac{a^2 + b^2}{a^2} \) i.e., \( \csc^2 51^\circ - 1 = \cfrac{a^2 + b^2}{a^2} - 1 \) i.e., \( \cot^2 51^\circ = \cfrac{\cancel{a^2} + b^2 - \cancel{a^2}}{a^2} \) i.e., \( \cot 51^\circ = \cfrac{b}{a} \) i.e., \( \tan 51^\circ = \cfrac{a}{b} \) \(\therefore \tan 51^\circ + \tan 39^\circ \) \(= \tan 51^\circ + \tan(90^\circ - 51^\circ) \) \(= \tan 51^\circ + \cot 51^\circ \) \(= \cfrac{a}{b} + \cfrac{b}{a} \) \(= \cfrac{a^2 + b^2}{ab} \)