A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{2}{\sqrt{2}}\]
C) \[2\sqrt{2}\]
D) \[\frac{\sqrt{3}}{2}\]
Correct Answer: C
Solution :
\[\text{cosec}\,A\,.\sec A\] \[=\frac{\sqrt{1+{{\tan }^{2}}A}}{\tan A}\cdot \sqrt{1+{{\tan }^{2}}A}\] \[=\frac{1+{{\tan }^{2}}A}{\tan A}=\frac{1+{{(\sqrt{2}-1)}^{2}}}{\sqrt{2}-1}\] \[=\frac{1+2+1-2\sqrt{2}}{\sqrt{2}-1}\] \[=\frac{4-2\sqrt{2}}{\sqrt{2}-1}=\frac{2\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}=2\sqrt{2}\]
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