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AP EAPCET 08-Jul-2022 Shift 1 Paper

Section: Mathematics

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Question : 69 of 160

Marks: +1, -0

The minimum value of

\left(1+\frac{1}{\sin^{n}\alpha}\right)\left(1+\frac{1}{\cos^{n}\alpha}\right)

is is

$1$
$2$
$(1+2^{n})^{2}$
$(1+2^{n/2})^{2}$

Solution: 👈: Video Solution

Let

y = \left(1+\frac{1}{\sin^{n}\alpha}\right)\left(1+\frac{1}{\cos^{n}\alpha}\right)

y = 1 + \frac{1}{\sin^{n}\alpha} + \frac{1}{\cos^{n}\alpha} + \frac{1}{\sin^{n}\alpha \cos^{n}\alpha}

y = 1 + \sin^{-n}\alpha + \cos^{-n}\alpha + \sin^{-n}\alpha \cos^{-n}\alpha

\frac{dy}{d\alpha} = \frac{-n}{\sin^{n+1}\alpha} \cos \alpha + \frac{n}{\cos^{n+1}\alpha} \sin \alpha + \frac{1}{\sin^{n}\alpha} \left[ \frac{n}{\sin^{n+1}\alpha} \sin \alpha \right]

+ \frac{1}{\cos^{n}\alpha} \left[ \frac{-n}{\sin^{n+1}\alpha} \cos \alpha \right]

= \frac{-n \cos \alpha}{\sin^{n+1}\alpha} \left(1+\frac{1}{\cos^{n}\alpha}\right) + \frac{n \sin \alpha}{\cos^{n+1}\alpha} \left(1+\frac{1}{\sin^{n}\alpha}\right)

\frac{dy}{dx} = 0, \text{ where } \alpha = \frac{\pi}{4}

y

at

\alpha = \frac{\pi}{4} = (1+\sqrt{2^n})^2 (1+\sqrt{2})^2

= (1+\sqrt{2^n})^2 = (1+2^{n/2})^2

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