If \[\tan x=\frac{b}{a},\]then \[\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\] [MP PET 1990, 2002]

\[\frac{2\cos x}{\sqrt{\sin 2x}}\]

\[\frac{2\sin x}{\sqrt{\sin 2x}}\]

\[\frac{2\cos x}{\sqrt{\cos 2x}}\]

\[\frac{2\sin x}{\sqrt{\cos 2x}}\]

If \[\tan x=\frac{b}{a},\]then \[\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-

1.	If \[\tan x=\frac{b}{a},\]then \[\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\] [MP PET 1990, 2002]
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.	\[\frac{2\sin x}{\sqrt{\cos 2x}}\]
Answer» C. \[\frac{2\cos x}{\sqrt{\sin 2x}}\]

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If \[\tan x=\frac{b}{a},\]then \[\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\] [MP PET 1990, 2002]

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