Find the value of tan-1⁡(13)+tan-1⁡(15)+tan-1⁡(\frac{1}{7})[/latex]

The correct option is tan⁻¹(7/9).

To find the value of the given expression, we use the formula for the sum of two inverse tangent functions:
$${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$$
provided that $xy<1$.

Let the given expression be:
$$S={\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$$

Step 1: Combine the first two terms
We apply the sum formula to ${\tan }^{-1}\left(\frac{1}{3}\right)$ and ${\tan }^{-1}\left(\frac{1}{5}\right)$:
$${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)={\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\left(\frac{1}{3}\right)\left(\frac{1}{5}\right)}\right)$$

Simplify the numerator and denominator:
Numerator: $\frac{1}{3}+\frac{1}{5}=\frac{5+3}{15}=\frac{8}{15}$
Denominator: $1-\frac{1}{15}=\frac{14}{15}$
Substituting these values back in:
$${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)={\tan }^{-1}\left(\frac{8/15}{14/15}\right)={\tan }^{-1}\left(\frac{8}{14}\right)={\tan }^{-1}\left(\frac{4}{7}\right)$$

Step 2: Add the third term to the result
Now, we add the third term ${\tan }^{-1}\left(\frac{1}{7}\right)$ to our simplified sum:
$$S={\tan }^{-1}\left(\frac{4}{7}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$$

Applying the sum formula again:
$$S={\tan }^{-1}\left(\frac{\frac{4}{7}+\frac{1}{7}}{1-\left(\frac{4}{7}\right)\left(\frac{1}{7}\right)}\right)$$

Simplify the fraction:
Numerator: $\frac{4}{7}+\frac{1}{7}=\frac{5}{7}$
Denominator: $1-\frac{4}{49}=\frac{49-4}{49}=\frac{45}{49}$
Substituting these back into the expression:
$$S={\tan }^{-1}\left(\frac{5/7}{45/49}\right)={\tan }^{-1}(\frac{5}{7}\times \frac{49}{45})$$

We simplify the product inside the argument:
$$\frac{5\times 49}{7\times 45}=\frac{49}{7\times 9}=\frac{7}{9}$$

Therefore, the simplified value is:
$$S={\tan }^{-1}\left(\frac{7}{9}\right)$$