If the mapping `f(x) = mx + c, m gt 0 ` maps `[-1,1]` onto `[0,2]`, then `tan ( tan^(-1). 1/7 + cot ^(-1) 8 + cot ^(-1) 18)` is equal to

Options

A. `f(2/3)`
B. `f(1/3)`
C. `f((-1)/3)`
D. `f((-2)/3)`

Answer

To solve the problem, we need to evaluate the expression \( \tan \left( \tan^{-1} \frac{1}{7} + \cot^{-1} 8 + \cot^{-1} 18 \right) \).

### Step 1: Rewrite the cotangent inverse terms
We can rewrite \( \cot^{-1} 8 \) and \( \cot^{-1} 18 \) in terms of tangent inverse:
\[ \cot^{-1} 8 = \tan^{-1} \frac{1}{8} \]
\[ \cot^{-1} 18 = \tan^{-1} \frac{1}{18} \]

### Step 2: Use the tangent addition formula
Now we can use the formula for the tangent of the sum of two angles:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Let \( A = \tan^{-1} \frac{1}{7} \) and \( B = \tan^{-1} \frac{1}{8} \).

### Step 3: Apply the formula
Using the formula:
\[ \tan \left( \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} \right) = \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \cdot \frac{1}{8}} \]

### Step 4: Simplify the expression
Calculate \( \frac{1}{7} + \frac{1}{8} \):
\[ \frac{1}{7} + \frac{1}{8} = \frac{8 + 7}{56} = \frac{15}{56} \]
Calculate \( 1 - \frac{1}{7} \cdot \frac{1}{8} \):
\[ 1 - \frac{1}{56} = \frac{56 - 1}{56} = \frac{55}{56} \]

### Step 5: Substitute back into the tangent formula
Now substituting back:
\[ \tan \left( \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} \right) = \frac{\frac{15}{56}}{\frac{55}{56}} = \frac{15}{55} = \frac{3}{11} \]

### Step 6: Add the last term
Now we need to add \( \tan^{-1} \frac{3}{11} \) and \( \tan^{-1} \frac{1}{18} \):
Using the tangent addition formula again:
\[ \tan \left( \tan^{-1} \frac{3}{11} + \tan^{-1} \frac{1}{18} \right) = \frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11} \cdot \frac{1}{18}} \]

### Step 7: Simplify this expression
Calculate \( \frac{3}{11} + \frac{1}{18} \):
\[ \frac{3 \cdot 18 + 1 \cdot 11}{11 \cdot 18} = \frac{54 + 11}{198} = \frac{65}{198} \]
Calculate \( 1 - \frac{3}{11} \cdot \frac{1}{18} \):
\[ 1 - \frac{3}{198} = \frac{198 - 3}{198} = \frac{195}{198} \]

### Step 8: Final substitution
Now substituting back:
\[ \tan \left( \tan^{-1} \frac{3}{11} + \tan^{-1} \frac{1}{18} \right) = \frac{\frac{65}{198}}{\frac{195}{198}} = \frac{65}{195} = \frac{13}{39} = \frac{1}{3} \]

### Final Result
Thus, we have:
\[ \tan \left( \tan^{-1} \frac{1}{7} + \cot^{-1} 8 + \cot^{-1} 18 \right) = \frac{1}{3} \]