If `|tanA|lt1and |A|`is acute, then
`(sqrt(1+sin2A)+sqrt(1-sin2A))/(sqrt(1+sin2A)-sqrt(1-sin2A))` is equal to

Options

A. `tan A`
B. `-tan A`
C. `cot A`
D. `-cot A`

Explanation

To solve the given expression

\[ \frac{\sqrt{1 + \sin 2A} + \sqrt{1 - \sin 2A}}{\sqrt{1 + \sin 2A} - \sqrt{1 - \sin 2A}}, \]

we will follow these steps:

### Step 1: Simplify the expression using trigonometric identities
We know that \(\sin 2A = 2 \sin A \cos A\). Thus, we can rewrite the expression as:

\[ \frac{\sqrt{1 + 2 \sin A \cos A} + \sqrt{1 - 2 \sin A \cos A}}{\sqrt{1 + 2 \sin A \cos A} - \sqrt{1 - 2 \sin A \cos A}}. \]

### Step 2: Recognize the form of the expression
The expression can be simplified further by recognizing that \(1 + \sin 2A\) and \(1 - \sin 2A\) can be expressed in terms of squares:

\[ 1 + \sin 2A = \cos^2 A + \sin^2 A + 2 \sin A \cos A = (\sin A + \cos A)^2, \]
\[ 1 - \sin 2A = \cos^2 A + \sin^2 A - 2 \sin A \cos A = (\cos A - \sin A)^2. \]

### Step 3: Substitute back into the expression
Now substituting these back into our expression gives:

\[ \frac{\sqrt{(\sin A + \cos A)^2} + \sqrt{(\cos A - \sin A)^2}}{\sqrt{(\sin A + \cos A)^2} - \sqrt{(\cos A - \sin A)^2}}. \]

### Step 4: Simplify the square roots
Since \(|\sin A + \cos A| = \sin A + \cos A\) and \(|\cos A - \sin A| = \cos A - \sin A\) (because both A is acute and \(|\tan A| < 1\)), we can simplify to:

\[ \frac{\sin A + \cos A + \cos A - \sin A}{\sin A + \cos A - (\cos A - \sin A)}. \]

### Step 5: Combine like terms
This simplifies to:

\[ \frac{2 \cos A}{2 \sin A} = \frac{\cos A}{\sin A} = \cot A. \]

### Final Result
Thus, the final result is:

\[ \cot A. \]