` 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 will follow these steps:

### Step 1: Substitute the Inverse Cosine
Let \( \theta = \cos^{-1}\left(\frac{a}{b}\right) \). Then, we have:
\[ \cos \theta = \frac{a}{b} \]

### Step 2: Rewrite the Expression
The expression can be rewritten as:
\[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) \]

### Step 3: Apply the Tan Addition and Subtraction Formulas
Using the formulas for tangent of a sum and difference:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
For \( a = \frac{\pi}{4} \) and \( b = \frac{\theta}{2} \), we know that \( \tan\left(\frac{\pi}{4}\right) = 1 \). Thus:
\[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{1 + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)} \]
\[ \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \frac{1 - \tan\left(\frac{\theta}{2}\right)}{1 + \tan\left(\frac{\theta}{2}\right)} \]

### Step 4: Combine the Two Tangent Expressions
Now, we combine the two results:
\[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \frac{1 + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)} + \frac{1 - \tan\left(\frac{\theta}{2}\right)}{1 + \tan\left(\frac{\theta}{2}\right)} \]

### Step 5: Simplify the Expression
Let \( x = \tan\left(\frac{\theta}{2}\right) \). Then, we have:
\[ \frac{(1 + x)(1 + x) + (1 - x)(1 - x)}{(1 - x)(1 + x)} = \frac{(1 + x)^2 + (1 - x)^2}{1 - x^2} \]
Expanding the numerator:
\[ (1 + 2x + x^2) + (1 - 2x + x^2) = 2 + 2x^2 \]
Thus, we have:
\[ \frac{2 + 2x^2}{1 - x^2} = \frac{2(1 + x^2)}{1 - x^2} \]

### Step 6: Use the Cosine Double Angle Identity
Recall that:
\[ \cos(2x) = \frac{1 - \tan^2 x}{1 + \tan^2 x} \]
So we can express this as:
\[ \frac{2}{\cos(\theta)} \]
Since \( \cos \theta = \frac{a}{b} \), we have:
\[ \frac{2}{\frac{a}{b}} = \frac{2b}{a} \]

### Final Answer
Thus, the final answer is:
\[ \frac{2b}{a} \]

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