Prove: `tan^(-1)(1/4)+tan^(-1)(2/9)=sin^(-1)(1/sqrt(5))`

Explanation

To prove that \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \), we will follow these steps:

### Step 1: Use the formula for the sum of inverse tangents
We start with the left-hand side:
\[ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \]
We can use the formula:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]
where \( x = \frac{1}{4} \) and \( y = \frac{2}{9} \).

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

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

### Step 3: Substitute into the formula
Now we substitute \( x + y \) and \( 1 - xy \) into the formula:
\[ \tan^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) = \tan^{-1}\left(\frac{17}{36} \cdot \frac{18}{17}\right) = \tan^{-1}\left(\frac{18}{36}\right) = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 4: Relate to sine inverse
Now we have:
\[ \tan^{-1}\left(\frac{1}{2}\right) \]
To convert this to sine inverse, we can consider a right triangle where the opposite side is 1 and the adjacent side is 2. The hypotenuse \( h \) is given by:
\[ h = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]
Thus, we can express the sine of the angle \( \theta \) as:
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}} \]
Therefore,
\[ \tan^{-1}\left(\frac{1}{2}\right) = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \]

### Conclusion
We have shown that:
\[ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \]
Thus, the left-hand side equals the right-hand side, proving the statement.