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The minimum value of $\left(1+\frac{1}{\sin ^n \alpha}\right)\left(1+\frac{1}{\cos ^n \alpha}\right)$ is
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Verified Answer
The correct answer is:
$\left(1+2^{n / 2}\right)^2$
Let $f(x)=\left(1+\frac{1}{\sin ^n x}\right)\left(1+\frac{1}{\cos ^n x}\right)$
$=\left(1+\operatorname{cosec}^n x\right)\left(1+\sec ^n x\right)$
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\left[-\mathrm{n} \operatorname{cosec}^{\mathrm{n}-1} \mathrm{x}-\operatorname{cosec} n \cdot \cot \mathrm{x}\right] \mathrm{dx} \\ & \left(1+\sec ^{\mathrm{n}} \mathrm{x}\right)+\left(1+\operatorname{cosec}^{\mathrm{n}} \mathrm{x}\right) \\ & \operatorname{nsec}^{\mathrm{n}} \mathrm{x} \times \sec \mathrm{x} \tan \mathrm{x}=0\end{aligned}$
$\Rightarrow \quad \mathrm{x}=\frac{\pi}{4}$
Now, $\mathrm{f}\left(\frac{\pi}{4}\right)=\left[1+(2)^{\mathrm{n} / 2}\right]^2$
$=\left(1+\operatorname{cosec}^n x\right)\left(1+\sec ^n x\right)$
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\left[-\mathrm{n} \operatorname{cosec}^{\mathrm{n}-1} \mathrm{x}-\operatorname{cosec} n \cdot \cot \mathrm{x}\right] \mathrm{dx} \\ & \left(1+\sec ^{\mathrm{n}} \mathrm{x}\right)+\left(1+\operatorname{cosec}^{\mathrm{n}} \mathrm{x}\right) \\ & \operatorname{nsec}^{\mathrm{n}} \mathrm{x} \times \sec \mathrm{x} \tan \mathrm{x}=0\end{aligned}$
$\Rightarrow \quad \mathrm{x}=\frac{\pi}{4}$
Now, $\mathrm{f}\left(\frac{\pi}{4}\right)=\left[1+(2)^{\mathrm{n} / 2}\right]^2$
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