If `tan alpha =(b)/(a), a gt b gt 0 and " if " 0 lt alpha lt (pi)/(4), " then " sqrt((a+b)/(a-b))+sqrt((a-b)/(a+b))` is equal to :

Options

A. `(2 sin alpha)/(sqrt(2 alpha))`
B. `(2 cos alpha)/(sqrt(cos2alpha))`
C. `(2 sin alpha)/(sqrt(sin 2alpha))`
D. `(2 cos alpha)/(sqrt(sin2alpha))`

Answer

To solve the problem step by step, we start with the given information and manipulate it to find the desired expression.

### Step 1: Start with the given expression
We are given that \( \tan \alpha = \frac{b}{a} \) where \( a > b > 0 \) and \( 0 < \alpha < \frac{\pi}{4} \). We need to evaluate:
\[ \sqrt{\frac{a+b}{a-b}} + \sqrt{\frac{a-b}{a+b}} \]

### Step 2: Rewrite the expression using \( \tan \alpha \)
We can express \( a \) and \( b \) in terms of \( \tan \alpha \):
Let \( b = a \tan \alpha \). Then, we can rewrite \( a + b \) and \( a - b \):
\[ a + b = a + a \tan \alpha = a(1 + \tan \alpha) \]
\[ a - b = a - a \tan \alpha = a(1 - \tan \alpha) \]

### Step 3: Substitute into the square root expressions
Now substituting these into our expression:
\[ \sqrt{\frac{a(1 + \tan \alpha)}{a(1 - \tan \alpha)}} + \sqrt{\frac{a(1 - \tan \alpha)}{a(1 + \tan \alpha)}} \]
This simplifies to:
\[ \sqrt{\frac{1 + \tan \alpha}{1 - \tan \alpha}} + \sqrt{\frac{1 - \tan \alpha}{1 + \tan \alpha}} \]

### Step 4: Simplify each term
Let \( x = \tan \alpha \). Then we have:
\[ \sqrt{\frac{1 + x}{1 - x}} + \sqrt{\frac{1 - x}{1 + x}} \]

### Step 5: Combine the fractions
We can combine these two square roots:
\[ \sqrt{\frac{1 + x}{1 - x}} + \sqrt{\frac{1 - x}{1 + x}} = \frac{(1+x) + (1-x)}{\sqrt{(1+x)(1-x)}} \]
This simplifies to:
\[ \frac{2}{\sqrt{1 - x^2}} \]

### Step 6: Substitute back for \( x \)
Recall that \( x = \tan \alpha \). Therefore, \( 1 - x^2 = 1 - \tan^2 \alpha = \cos^2 \alpha \) (using the identity \( \sec^2 \alpha = 1 + \tan^2 \alpha \)):
\[ \sqrt{1 - \tan^2 \alpha} = \sqrt{\cos^2 \alpha} = \cos \alpha \]

### Step 7: Final expression
Thus, we have:
\[ \frac{2}{\cos \alpha} = 2 \sec \alpha \]

### Conclusion
The final answer is:
\[ \sqrt{\frac{a+b}{a-b}} + \sqrt{\frac{a-b}{a+b}} = 2 \sec \alpha \]