A) \[\frac{2\sin x}{\sqrt{\sin 2x}}\]
B) \[\frac{2\cos x}{\sqrt{\cos 2x}}\]
C) \[\frac{2\cos x}{\sqrt{\sin 2x}}\]
D) \[\frac{2\sin x}{\sqrt{\cos 2x}}\]
Correct Answer: B
Solution :
Given that, \[\tan x=\frac{b}{a}\] Now \[\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\sqrt{\frac{1+b/a}{1-b/a}}+\sqrt{\frac{1-b/a}{1+b/a}}\] \[=\frac{2}{\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}}=\frac{2}{\sqrt{1-{{\tan }^{2}}x}}=\frac{2}{\sqrt{1-\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}}=\frac{2\cos x}{\sqrt{\cos 2x}}\].
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