If \(\tan^2 x = 1 - a^2\), then the value of \(\sec x + \tan^3 x \,\mathrm{cosec}\, x\) is:
\((2 - a^2)^{\frac{3}{2}}\)
Write \(\tan^3 x \,\mathrm{cosec}\, x = \tan^2 x \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\sin x} = \tan^2 x \sec x\).
So the expression becomes \(\sec x + \tan^2 x \sec x = \sec x (1 + \tan^2 x)\).
Using \(1 + \tan^2 x = \sec^2 x\), the expression equals \(\sec^3 x = (\sec^2 x)^{\frac{3}{2}} = (1 + \tan^2 x)^{\frac{3}{2}}\).
Substituting \(\tan^2 x = 1 - a^2\): \((1 + 1 - a^2)^{\frac{3}{2}} = (2 - a^2)^{\frac{3}{2}}\).
Hence, the answer is \((2 - a^2)^{\frac{3}{2}}\).
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