A) 6
B) 9
C) 18
D) 24
Correct Answer: C
Solution :
[c] \[\frac{1}{{{\sin }^{2}}\alpha }+\frac{1}{1+{{\cos }^{2}}\alpha }=\frac{1}{1-{{\cos }^{2}}\alpha }+\frac{1}{1+{{\cos }^{2}}\alpha }\] \[=\frac{2}{1-{{\cos }^{4}}\alpha }\] Now, \[\frac{2}{1-{{\cos }^{4}}\alpha }+\frac{2}{1+{{\cos }^{4}}\alpha }=\frac{4}{1-{{\cos }^{8}}\alpha }\] \[\therefore \,\,\,\frac{4}{1-{{\cos }^{8}}\alpha }+\frac{4}{1+{{\cos }^{8}}\alpha }=\frac{8}{1-{{\cos }^{16}}\alpha }\] and \[\frac{8}{1-{{\cos }^{16}}\alpha }+\frac{8}{1+{{\cos }^{16}}\alpha }=\frac{16}{1-{{\cos }^{32}}\alpha }=18\]
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