Let ABC be a right with `angleBAC=(pi)/(2)`, then `((r^(2))/(2R^(2))+(r )/(R ))` is equal to :
(where symbols used have usual meaning in a striangle)

Options

A. sinB sinC
B. tanB tanC
C. secB secC
D. cotB cotC

Answer

To solve the problem, we need to find the value of the expression \(\frac{r^2}{2R^2} + \frac{r}{R}\) for a right triangle \(ABC\) where \(\angle BAC = \frac{\pi}{2}\).

### Step-by-Step Solution:

1. **Identify the Triangle and Its Properties:**
- Given triangle \(ABC\) is a right triangle with \(\angle BAC = 90^\circ\).
- Let \(a\), \(b\), and \(c\) be the lengths of the sides opposite to angles \(A\), \(B\), and \(C\) respectively. Here, \(a\) is the hypotenuse.

2. **Calculate the Circumradius \(R\):**
- The formula for the circumradius \(R\) of a right triangle is given by:
\[ R = \frac{a}{2} \]
- Here, \(a\) is the length of the hypotenuse.

3. **Calculate the Inradius \(r\):**
- The inradius \(r\) can be calculated using the formula:
\[ r = \frac{A}{s} \]
where \(A\) is the area of the triangle and \(s\) is the semi-perimeter.
- The area \(A\) of triangle \(ABC\) is:
\[ A = \frac{1}{2} \times b \times c \]
- The semi-perimeter \(s\) is:
\[ s = \frac{a + b + c}{2} \]
- Therefore, the inradius \(r\) can also be expressed as:
\[ r = \frac{b + c - a}{2} \]

4. **Substitute \(R\) and \(r\) into the Expression:**
- We need to evaluate:
\[ \frac{r^2}{2R^2} + \frac{r}{R} \]
- Substitute \(R = \frac{a}{2}\) and \(r = \frac{b + c - a}{2}\):
\[ R^2 = \left(\frac{a}{2}\right)^2 = \frac{a^2}{4} \]
\[ r^2 = \left(\frac{b + c - a}{2}\right)^2 = \frac{(b + c - a)^2}{4} \]

5. **Calculate Each Term:**
- The first term becomes:
\[ \frac{r^2}{2R^2} = \frac{\frac{(b + c - a)^2}{4}}{2 \cdot \frac{a^2}{4}} = \frac{(b + c - a)^2}{2a^2} \]
- The second term becomes:
\[ \frac{r}{R} = \frac{\frac{b + c - a}{2}}{\frac{a}{2}} = \frac{b + c - a}{a} \]

6. **Combine the Terms:**
- Now combine both terms:
\[ \frac{(b + c - a)^2}{2a^2} + \frac{b + c - a}{a} \]
- To combine these, find a common denominator:
\[ = \frac{(b + c - a)^2 + 2a(b + c - a)}{2a^2} \]
- Simplifying the numerator:
\[ = \frac{(b + c - a)^2 + 2ab + 2ac - 2a^2}{2a^2} \]

7. **Final Simplification:**
- The expression simplifies to:
\[ = \frac{b^2 + c^2 + 2bc - a^2}{2a^2} \]
- Using the Pythagorean theorem \(a^2 = b^2 + c^2\):
\[ = \frac{2bc}{2a^2} = \frac{bc}{a^2} \]

### Final Result:
Thus, the value of \(\frac{r^2}{2R^2} + \frac{r}{R}\) is equal to:
\[ \frac{bc}{a^2} \]