Find the value of \(\frac{\tan A}{1 - \sqrt{\mathrm{cosec}^2 A - 1}} + \frac{\cot A}{1 - \sqrt{\sec^2 A - 1}}\).
\(1 + \tan A + \cot A\)
Use the identities \(\mathrm{cosec}^2 A - 1 = \cot^2 A\) and \(\sec^2 A - 1 = \tan^2 A\), so the roots become \(\cot A\) and \(\tan A\).
The expression becomes \(\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}\).
Let \(\tan A = t\), so \(\cot A = \frac{1}{t}\). Then \(\frac{t}{1 - \frac{1}{t}} = \frac{t^2}{t - 1}\) and \(\frac{\frac{1}{t}}{1 - t} = -\frac{1}{t(t - 1)}\).
Adding: \(\frac{t^3 - 1}{t(t - 1)} = \frac{(t - 1)(t^2 + t + 1)}{t(t - 1)} = \frac{t^2 + t + 1}{t}\).
\(\frac{t^2 + t + 1}{t} = t + 1 + \frac{1}{t} = 1 + \tan A + \cot A\).
Hence, the answer is \(1 + \tan A + \cot A\).
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