A) 1
B) 2
C) \[{{(1+{{2}^{n/2}})}^{2}}\]
D) none of these
Correct Answer: C
Solution :
Let \[f\,(\alpha )=\left( 1+\frac{1}{{{\sin }^{n}}\alpha } \right)\,\,\left( 1+\frac{1}{co{{s}^{n}}\alpha } \right)=1+\frac{1}{{{\sin }^{n}}\alpha }+\frac{1}{{{\cos }^{n}}\alpha }\]\[+\frac{1}{{{\sin }^{n}}\alpha {{\cos }^{n}}\alpha }\] Now, \[f'(\alpha )=0\] \[\Rightarrow \]\[\cos \alpha =\sin \alpha \]\[\Rightarrow \]\[\alpha =\pi /4.\] Hence \[f\,(\alpha )\] is maximum at \[\alpha =0\] and a \[\alpha =\pi /2\] & between two maxima, there is one minima. Hence \[\alpha =\pi /4\] gives the minimum value of \[f\,(\alpha )\] and is given by \[f\,(\pi /4)={{(1+{{2}^{n/2}})}^{2}}\]
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