If `cos theta=a/(b+c), cos phi= b/(a+c)` and `cos psi=c/(a+b)` where `theta, phi, psi in (0, pi)` and `a,b,c` are sides of triangle `ABC` then `tan^2(theta/2)+tan^2(phi/2)+tan^2(psi/2)=`

Explanation

To solve the problem, we need to find the value of \( \tan^2\left(\frac{\theta}{2}\right) + \tan^2\left(\frac{\phi}{2}\right) + \tan^2\left(\frac{\psi}{2}\right) \) given the expressions for \( \cos \theta \), \( \cos \phi \), and \( \cos \psi \).

### Step-by-Step Solution:

1. **Write down the given expressions:**
\[ \cos \theta = \frac{a}{b+c}, \quad \cos \phi = \frac{b}{a+c}, \quad \cos \psi = \frac{c}{a+b} \]

2. **Use the half-angle tangent formula:**
The formula for cosine in terms of tangent of half-angle is:
\[ \cos \theta = \frac{1 - \tan^2\left(\frac{\theta}{2}\right)}{1 + \tan^2\left(\frac{\theta}{2}\right)} \]
Rearranging gives:
\[ \frac{1 + \tan^2\left(\frac{\theta}{2}\right)}{1 - \tan^2\left(\frac{\theta}{2}\right)} = \frac{b+c}{a} \]

3. **Apply the component and dividendo method:**
Using the component and dividendo method:
\[ \tan^2\left(\frac{\theta}{2}\right) = \frac{b+c-a}{a+b+c} \]

4. **Repeat for \( \phi \) and \( \psi \):**
Similarly, we can find:
\[ \tan^2\left(\frac{\phi}{2}\right) = \frac{a+c-b}{a+b+c} \]
\[ \tan^2\left(\frac{\psi}{2}\right) = \frac{a+b-c}{a+b+c} \]

5. **Sum the expressions:**
Now, we can sum these three expressions:
\[ \tan^2\left(\frac{\theta}{2}\right) + \tan^2\left(\frac{\phi}{2}\right) + \tan^2\left(\frac{\psi}{2}\right) = \frac{(b+c-a) + (a+c-b) + (a+b-c)}{a+b+c} \]

6. **Simplify the numerator:**
Simplifying the numerator:
\[ (b+c-a) + (a+c-b) + (a+b-c) = a + b + c \]

7. **Final expression:**
Thus, we have:
\[ \tan^2\left(\frac{\theta}{2}\right) + \tan^2\left(\frac{\phi}{2}\right) + \tan^2\left(\frac{\psi}{2}\right) = \frac{a+b+c}{a+b+c} = 1 \]

### Conclusion:
The value of \( \tan^2\left(\frac{\theta}{2}\right) + \tan^2\left(\frac{\phi}{2}\right) + \tan^2\left(\frac{\psi}{2}\right) \) is \( 1 \).