`x=4tan^(-1)((1)/(5)),y=tan^(-1)((1)/(70))`
`z=tan^(-1)((1)/(99))`
What is x equal to?

Options

A. `tan^(-1)((60)/(119))`
B. `tan^(-1)((120)/(119))`
C. `tan^(-1)((90)/(169))`
D. `tan^(-1)((170)/(169))`

Answer

To solve for \( x = 4 \tan^{-1} \left( \frac{1}{5} \right) \), we can use the double angle formula for the tangent inverse function. The formula states:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \quad \text{if } ab < 1 \]

### Step 1: Find \( 2 \tan^{-1} \left( \frac{1}{5} \right) \)

Using the double angle formula:

\[ 2 \tan^{-1}(x) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]

Let \( x = \frac{1}{5} \):

\[ 2 \tan^{-1} \left( \frac{1}{5} \right) = \tan^{-1} \left( \frac{2 \cdot \frac{1}{5}}{1 - \left( \frac{1}{5} \right)^2} \right) \]

Calculating the numerator:

\[ 2 \cdot \frac{1}{5} = \frac{2}{5} \]

Calculating the denominator:

\[ 1 - \left( \frac{1}{5} \right)^2 = 1 - \frac{1}{25} = \frac{25 - 1}{25} = \frac{24}{25} \]

So we have:

\[ 2 \tan^{-1} \left( \frac{1}{5} \right) = \tan^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} \right) = \tan^{-1} \left( \frac{2 \cdot 25}{5 \cdot 24} \right) = \tan^{-1} \left( \frac{10}{24} \right) = \tan^{-1} \left( \frac{5}{12} \right) \]

### Step 2: Find \( 4 \tan^{-1} \left( \frac{1}{5} \right) \)

Now we can apply the double angle formula again to find \( 4 \tan^{-1} \left( \frac{1}{5} \right) \):

\[ 4 \tan^{-1}(x) = 2 \tan^{-1}(2x/(1-x^2)) \]

Letting \( x = \frac{5}{12} \):

\[ 4 \tan^{-1} \left( \frac{1}{5} \right) = 2 \tan^{-1} \left( \frac{2 \cdot \frac{5}{12}}{1 - \left( \frac{5}{12} \right)^2} \right) \]

Calculating the numerator:

\[ 2 \cdot \frac{5}{12} = \frac{10}{12} = \frac{5}{6} \]

Calculating the denominator:

\[ 1 - \left( \frac{5}{12} \right)^2 = 1 - \frac{25}{144} = \frac{144 - 25}{144} = \frac{119}{144} \]

So we have:

\[ 4 \tan^{-1} \left( \frac{1}{5} \right) = 2 \tan^{-1} \left( \frac{\frac{5}{6}}{\frac{119}{144}} \right) = 2 \tan^{-1} \left( \frac{5 \cdot 144}{6 \cdot 119} \right) = 2 \tan^{-1} \left( \frac{120}{119} \right) \]

### Step 3: Final Calculation

Now we apply the double angle formula one more time:

\[ 2 \tan^{-1}(x) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]

Let \( x = \frac{120}{119} \):

\[ \tan^{-1} \left( \frac{2 \cdot \frac{120}{119}}{1 - \left( \frac{120}{119} \right)^2} \right) \]

Calculating the numerator:

\[ 2 \cdot \frac{120}{119} = \frac{240}{119} \]

Calculating the denominator:

\[ 1 - \left( \frac{120}{119} \right)^2 = 1 - \frac{14400}{14161} = \frac{14161 - 14400}{14161} = \frac{-239}{14161} \]

So we have:

\[ x = \tan^{-1} \left( \frac{\frac{240}{119}}{\frac{-239}{14161}} \right) = \tan^{-1} \left( \frac{240 \cdot 14161}{-239 \cdot 119} \right) \]

Thus, the value of \( x \) is:

\[ x = \tan^{-1} \left( \frac{120}{119} \right) \]

### Final Answer

\[ x = \tan^{-1} \left( \frac{120}{119} \right) \]