Discussion
If tan2 α= 1-e2, then the value of sec α + tan3 α. Cosec α is :
Answer: D
The value is \((2 - e^{2})^{3/2}\).
Reading the expression as \(\sec\alpha + \tan^{3}\alpha \cdot \csc\alpha\), first simplify the second term:
\[ \begin{aligned} \tan^{3}\alpha \csc\alpha &= \tan^{2}\alpha \cdot \frac{\sin\alpha}{\cos\alpha} \cdot \frac{1}{\sin\alpha} \\ &= \tan^{2}\alpha \sec\alpha \end{aligned} \]
so the whole expression is
\[ \begin{aligned} \sec\alpha(1 + \tan^{2}\alpha) &= \sec^{3}\alpha \\ &= (1 + \tan^{2}\alpha)^{3/2} \\ &= (2 - e^{2})^{3/2} \end{aligned} \]
\((2 - e^{2})^{1/2}\) is \(\sec\alpha\) alone, the first term without the second. The options with \(2 + e^{2}\) get the sign of \(e^{2}\) wrong when adding 1 to \(1 - e^{2}\).
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