`tan^(- 1)(1/4)+tan^(- 1)(2/9)=1/2tan^(- 1)(4/3)`

Options

A. `1/2cos^(-1)(3/5)`
B. `1/2sin^(-1)(3/5)`
C. `1/2tan^(-1)(3/5)`
D. `tan^(-1)(1/2)`

Answer

To prove that \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \frac{1}{2} \tan^{-1}\left(\frac{4}{3}\right) \), we will follow these steps:

### Step 1: Write the Left-Hand Side
We start with the left-hand side of the equation:
\[ LHS = \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \]

### Step 2: Use the Formula for the Sum of Inverses
We use the formula for the sum of two inverse tangents:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
Here, let \( x = \frac{1}{4} \) and \( y = \frac{2}{9} \).

### Step 3: Calculate \( x + y \) and \( 1 - xy \)
Calculate \( x + y \):
\[ x + y = \frac{1}{4} + \frac{2}{9} \]
To add these fractions, we find a common denominator, which is 36:
\[ x + y = \frac{9}{36} + \frac{8}{36} = \frac{17}{36} \]

Now calculate \( 1 - xy \):
\[ xy = \frac{1}{4} \cdot \frac{2}{9} = \frac{2}{36} = \frac{1}{18} \]
Thus,
\[ 1 - xy = 1 - \frac{1}{18} = \frac{18}{18} - \frac{1}{18} = \frac{17}{18} \]

### Step 4: Substitute Back into the Formula
Now substitute back into the formula:
\[ LHS = \tan^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) \]
This simplifies to:
\[ LHS = \tan^{-1}\left(\frac{17}{36} \cdot \frac{18}{17}\right) = \tan^{-1}\left(\frac{18}{36}\right) = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 5: Relate to Right-Hand Side
Now, we need to show that:
\[ \tan^{-1}\left(\frac{1}{2}\right) = \frac{1}{2} \tan^{-1}\left(\frac{4}{3}\right) \]
Using the formula \( 2 \tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \), we set \( x = \frac{1}{2} \):
\[ 2 \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2}\right) = \tan^{-1}\left(\frac{1}{1 - \frac{1}{4}}\right) = \tan^{-1}\left(\frac{1}{\frac{3}{4}}\right) = \tan^{-1}\left(\frac{4}{3}\right) \]
Thus,
\[ \tan^{-1}\left(\frac{1}{2}\right) = \frac{1}{2} \tan^{-1}\left(\frac{4}{3}\right) \]

### Conclusion
We have shown that:
\[ LHS = \frac{1}{2} \tan^{-1}\left(\frac{4}{3}\right) = RHS \]
Therefore, the equation is proven.