If \(\cos A = \frac{4}{5}\), find the value of \(\dfrac{\sin^2 A - 5\cos^2 A + \sec^2 A}{\tan^2 A}\).
\(\frac{-511}{225}\)
Given \(\cos A = \frac{4}{5}\), so \(\sin A = \sqrt{1-\frac{16}{25}} = \frac{3}{5}\).
Then \(\tan A = \frac{\sin A}{\cos A} = \frac{3}{4}\) and \(\sec A = \frac{1}{\cos A} = \frac{5}{4}\).
Numerator \(= \sin^2 A - 5\cos^2 A + \sec^2 A = \frac{9}{25} - 5\cdot\frac{16}{25} + \frac{25}{16}\).
\(= \frac{9}{25} - \frac{80}{25} + \frac{25}{16} = -\frac{71}{25} + \frac{25}{16} = \frac{-1136 + 625}{400} = -\frac{511}{400}\).
Denominator \(= \tan^2 A = \frac{9}{16}\).
\(\dfrac{-\frac{511}{400}}{\frac{9}{16}} = -\frac{511}{400} \times \frac{16}{9} = -\frac{511}{225}\).
Hence, the value is \(-\frac{511}{225}\).
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