In a triangle ABC, if `r_1 + r_3 + r = r_2`, then find the value of `(sec^2 A + csc^2 B-cot^2 C)`,

Answer

To solve the problem step by step, we start with the given condition and proceed through the necessary mathematical deductions.

### Step 1: Understand the Given Condition
We are given that in triangle ABC, \( r_1 + r_3 + r = r_2 \). Here, \( r_1, r_2, r_3 \) are the exradii of the triangle opposite to vertices A, B, and C respectively, and \( r \) is the inradius.

### Step 2: Express the Exradii in Terms of Area and Semiperimeter
The exradii can be expressed as:
- \( r = \frac{\Delta}{s} \)
- \( r_1 = \frac{\Delta}{s - a} \)
- \( r_2 = \frac{\Delta}{s - b} \)
- \( r_3 = \frac{\Delta}{s - c} \)

Where \( \Delta \) is the area of the triangle and \( s \) is the semiperimeter given by \( s = \frac{a + b + c}{2} \).

### Step 3: Substitute the Exradii into the Given Condition
Substituting these expressions into the given condition:
\[ \frac{\Delta}{s - a} + \frac{\Delta}{s - c} + \frac{\Delta}{s} = \frac{\Delta}{s - b} \]

### Step 4: Simplify the Equation
We can factor out \( \Delta \) (assuming \( \Delta \neq 0 \)):
\[ \frac{1}{s - a} + \frac{1}{s - c} + \frac{1}{s} = \frac{1}{s - b} \]

### Step 5: Find a Common Denominator
The common denominator for the left-hand side is \( (s - a)(s - c)s \):
\[ \frac{s(s - c) + s(s - a) + (s - a)(s - c)}{(s - a)(s - c)s} = \frac{(s - b)(s - a)(s - c)}{(s - b)(s - a)(s - c)} \]

### Step 6: Expand and Rearrange
After expanding and simplifying, we will reach the conclusion that:
\[ a^2 + c^2 = b^2 \]
This indicates that triangle ABC is a right triangle with the right angle at B.

### Step 7: Use the Right Triangle Properties
In a right triangle at B, we can use the relationships:
- \( \sec^2 A = 1 + \tan^2 A \)
- \( \csc^2 B = 1 \) (since \( \sin B = 1 \))
- \( \cot^2 C = \frac{b^2}{a^2} \)

### Step 8: Calculate the Required Expression
Now we need to find:
\[ \sec^2 A + \csc^2 B - \cot^2 C \]
Substituting the values:
\[ \sec^2 A + 1 - \frac{b^2}{a^2} \]

Using \( b^2 = a^2 + c^2 \):
\[ = \sec^2 A + 1 - \frac{a^2 + c^2}{a^2} \]
This simplifies to:
\[ = \sec^2 A + 1 - 1 - \frac{c^2}{a^2} \]
Thus, we have:
\[ = \sec^2 A - \frac{c^2}{a^2} \]

### Step 9: Final Calculation
Since \( \sec^2 A = \frac{c^2}{b^2} \) in a right triangle:
\[ = 1 \]

### Conclusion
The value of \( \sec^2 A + \csc^2 B - \cot^2 C \) is \( 1 \).

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