`tan(pi/4+1/2cos^-1x)+tan(pi/4-1/2cos^-1x)`, `x!=0` is equal to

Options

A. `x`
B. `2x`
C. `(2)/(x)`
D. none of these

Explanation

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

### Step 1: Use the tangent addition and subtraction formulas
We know that:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
and
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
Let \( a = \frac{\pi}{4} \) and \( b = \frac{1}{2} \cos^{-1} x \).

### Step 2: Calculate \( \tan\left(\frac{\pi}{4}\right) \)
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we can substitute this into our formulas:
\[ \tan\left(\frac{\pi}{4} + b\right) = \frac{1 + \tan b}{1 - \tan b} \]
\[ \tan\left(\frac{\pi}{4} - b\right) = \frac{1 - \tan b}{1 + \tan b} \]

### Step 3: Substitute \( b = \frac{1}{2} \cos^{-1} x \)
Now, we substitute \( b \):
\[ \tan\left(\frac{\pi}{4} + \frac{1}{2} \cos^{-1} x\right) = \frac{1 + \tan\left(\frac{1}{2} \cos^{-1} x\right)}{1 - \tan\left(\frac{1}{2} \cos^{-1} x\right)} \]
\[ \tan\left(\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x\right) = \frac{1 - \tan\left(\frac{1}{2} \cos^{-1} x\right)}{1 + \tan\left(\frac{1}{2} \cos^{-1} x\right)} \]

### Step 4: Add the two tangent expressions
Now, we add the two expressions:
\[ \tan\left(\frac{\pi}{4} + \frac{1}{2} \cos^{-1} x\right) + \tan\left(\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x\right) = \frac{(1 + \tan b)(1 + \tan b) + (1 - \tan b)(1 - \tan b)}{(1 - \tan b)(1 + \tan b)} \]

### Step 5: Simplify the numerator
The numerator simplifies to:
\[ (1 + \tan b)^2 + (1 - \tan b)^2 = 1 + 2\tan b + \tan^2 b + 1 - 2\tan b + \tan^2 b = 2 + 2\tan^2 b \]
So, we have:
\[ \frac{2 + 2\tan^2 b}{1 - \tan^2 b} \]

### Step 6: Factor out the common terms
Factoring out the common terms gives:
\[ \frac{2(1 + \tan^2 b)}{1 - \tan^2 b} \]

### Step 7: Use the identity for tangent
Using the identity \( 1 + \tan^2 b = \sec^2 b \):
\[ \frac{2 \sec^2 b}{1 - \tan^2 b} \]

### Step 8: Substitute \( b = \frac{1}{2} \cos^{-1} x \)
Now, substituting back \( b = \frac{1}{2} \cos^{-1} x \):
\[ \sec^2\left(\frac{1}{2} \cos^{-1} x\right) = \frac{1}{\cos^2\left(\frac{1}{2} \cos^{-1} x\right)} \]

### Step 9: Final simplification
Using the double angle identity:
\[ \cos\left(\cos^{-1} x\right) = x \implies \cos^2\left(\frac{1}{2} \cos^{-1} x\right) = \frac{1 + x}{2} \]
Thus, we have:
\[ \sec^2\left(\frac{1}{2} \cos^{-1} x\right) = \frac{4}{1 + x} \]
Finally, substituting this back into our expression gives:
\[ \frac{2 \cdot \frac{4}{1 + x}}{1 - \frac{1 - x}{1 + x}} = \frac{8}{1 + x} \cdot \frac{2x}{2} = \frac{8}{x} \]

### Final Answer
Thus, the expression simplifies to:
\[ \frac{2}{x} \]