Answer: A
Given: \(\sin 51^\circ = \cfrac{a}{\sqrt{a^2 + b^2}}\) Then, \(\cosec 51^\circ = \cfrac{\sqrt{a^2 + b^2}}{a}\) \(\Rightarrow \cosec^2 51^\circ = \cfrac{a^2 + b^2}{a^2}\) \(\Rightarrow 1 + \cot^2 51^\circ = \cfrac{a^2 + b^2}{a^2}\) \(\Rightarrow \cot^2 51^\circ = \cfrac{a^2 + b^2}{a^2} - 1\) \(\Rightarrow \cot^2 51^\circ = \cfrac{a^2 + b^2 - a^2}{a^2} = \cfrac{b^2}{a^2}\) \(\Rightarrow \cot 51^\circ = \cfrac{b}{a}\) Now, \(\tan 51^\circ + \tan 39^\circ\) \(= \cfrac{1}{\cot 51^\circ} + \tan(90^\circ - 51^\circ)\) \(= \cfrac{1}{\cot 51^\circ} + \cot 51^\circ\) \(= \cfrac{1}{\cfrac{b}{a}} + \cfrac{b}{a}\) \(= \cfrac{a}{b} + \cfrac{b}{a}\) \(= \cfrac{a^2 + b^2}{ab}\)
Given: \(\sin 51^\circ = \cfrac{a}{\sqrt{a^2 + b^2}}\) Then, \(\cosec 51^\circ = \cfrac{\sqrt{a^2 + b^2}}{a}\) \(\Rightarrow \cosec^2 51^\circ = \cfrac{a^2 + b^2}{a^2}\) \(\Rightarrow 1 + \cot^2 51^\circ = \cfrac{a^2 + b^2}{a^2}\) \(\Rightarrow \cot^2 51^\circ = \cfrac{a^2 + b^2}{a^2} - 1\) \(\Rightarrow \cot^2 51^\circ = \cfrac{a^2 + b^2 - a^2}{a^2} = \cfrac{b^2}{a^2}\) \(\Rightarrow \cot 51^\circ = \cfrac{b}{a}\) Now, \(\tan 51^\circ + \tan 39^\circ\) \(= \cfrac{1}{\cot 51^\circ} + \tan(90^\circ - 51^\circ)\) \(= \cfrac{1}{\cot 51^\circ} + \cot 51^\circ\) \(= \cfrac{1}{\cfrac{b}{a}} + \cfrac{b}{a}\) \(= \cfrac{a}{b} + \cfrac{b}{a}\) \(= \cfrac{a^2 + b^2}{ab}\)