` tan((pi)/(4) +(1)/(2)"cos"^(-1)(a)/(b))+ tan((pi)/(4) - (1)/(2)"cos"^(-1)(a)/(b)) ` is :

Options

A. `(2a)/(b) `
B. ` (a)/(b) `
C. ` (b)/(a) `
D. `(2b)/(a) `

Answer

To solve the expression \( \tan\left(\frac{\pi}{4} + \frac{1}{2} \cos^{-1}\left(\frac{a}{b}\right)\right) + \tan\left(\frac{\pi}{4} - \frac{1}{2} \cos^{-1}\left(\frac{a}{b}\right)\right) \), we can follow these steps:

### Step 1: Define the angle
Let:
\[ \theta = \frac{1}{2} \cos^{-1}\left(\frac{a}{b}\right) \]
Now, we can rewrite the expression as:
\[ \tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) \]

### Step 2: Use the tangent addition and subtraction formulas
Using the tangent addition formula:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
and the tangent subtraction formula:
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
we can apply these to our expression. Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we have:
\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan \theta}{1 - 1 \cdot \tan \theta} = \frac{1 + \tan \theta}{1 - \tan \theta} \]
\[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + 1 \cdot \tan \theta} = \frac{1 - \tan \theta}{1 + \tan \theta} \]

### Step 3: Combine the two expressions
Now we can combine these two results:
\[ \tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 + \tan \theta}{1 - \tan \theta} + \frac{1 - \tan \theta}{1 + \tan \theta} \]

### Step 4: Find a common denominator
The common denominator for these two fractions is \( (1 - \tan \theta)(1 + \tan \theta) \):
\[ = \frac{(1 + \tan \theta)^2 + (1 - \tan \theta)^2}{(1 - \tan \theta)(1 + \tan \theta)} \]

### Step 5: Simplify the numerator
Now we simplify the numerator:
\[ (1 + \tan \theta)^2 + (1 - \tan \theta)^2 = 1 + 2\tan \theta + \tan^2 \theta + 1 - 2\tan \theta + \tan^2 \theta = 2 + 2\tan^2 \theta \]

### Step 6: Simplify the denominator
The denominator can be simplified as:
\[ (1 - \tan \theta)(1 + \tan \theta) = 1 - \tan^2 \theta \]

### Step 7: Final expression
Putting it all together, we have:
\[ \tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) = \frac{2 + 2\tan^2 \theta}{1 - \tan^2 \theta} \]
Factoring out a 2 from the numerator gives:
\[ = \frac{2(1 + \tan^2 \theta)}{1 - \tan^2 \theta} \]

### Step 8: Use the double angle identity for cosine
We know that:
\[ 1 + \tan^2 \theta = \sec^2 \theta \]
and using the double angle identity for cosine:
\[ \cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
Thus, we can express the final answer as:
\[ = \frac{2}{\cos(2\theta)} \]
Substituting back \( \theta = \frac{1}{2} \cos^{-1}\left(\frac{a}{b}\right) \):
\[ = \frac{2}{\cos\left(\cos^{-1}\left(\frac{a}{b}\right)\right)} = \frac{2b}{a} \]

### Final Answer
The final result is:
\[ \frac{2b}{a} \]