If \(\sin A = \dfrac{m}{n}\), then what is the value of \((1 + \tan^2 A)\)?
\(\dfrac{n^2}{n^2-m^2}\)
\(1 + \tan^2 A = \sec^2 A = \dfrac{1}{\cos^2 A}\)
\(\cos^2 A = 1 - \sin^2 A = 1 - \dfrac{m^2}{n^2} = \dfrac{n^2 - m^2}{n^2}\)
\(\sec^2 A = \dfrac{n^2}{n^2 - m^2}\)
Therefore \(1 + \tan^2 A = \mathbf{\dfrac{n^2}{n^2-m^2}}\)
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