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 \(ABC\) with \(\angle BAC = 90^\circ\).

### Step-by-Step Solution:

1. **Identify the Triangle Properties**:
In a right triangle, the circumradius \(R\) and the inradius \(r\) can be expressed in terms of the sides of the triangle. Let \(a\), \(b\), and \(c\) be the lengths of the sides opposite to angles \(A\), \(B\), and \(C\) respectively. Since \(\angle A = 90^\circ\), we have:
- \(R = \frac{a}{2}\)
- \(r = \frac{b + c - a}{2}\)

2. **Substituting Values**:
Now, substituting \(R\) and \(r\) into the expression:
\[ \frac{r^2}{2R^2} + \frac{r}{R} \]
becomes:
\[ \frac{\left(\frac{b+c-a}{2}\right)^2}{2\left(\frac{a}{2}\right)^2} + \frac{\frac{b+c-a}{2}}{\frac{a}{2}} \]

3. **Simplifying the First Term**:
The first term simplifies as follows:
\[ \frac{\left(\frac{b+c-a}{2}\right)^2}{2\left(\frac{a^2}{4}\right)} = \frac{(b+c-a)^2}{2 \cdot \frac{a^2}{4}} = \frac{(b+c-a)^2}{\frac{a^2}{2}} = \frac{2(b+c-a)^2}{a^2} \]

4. **Simplifying the Second Term**:
The second term simplifies as:
\[ \frac{\frac{b+c-a}{2}}{\frac{a}{2}} = \frac{b+c-a}{a} \]

5. **Combining the Terms**:
Now, we combine both terms:
\[ \frac{2(b+c-a)^2}{a^2} + \frac{b+c-a}{a} \]
To combine these, we need a common denominator, which is \(a^2\):
\[ \frac{2(b+c-a)^2 + (b+c-a)a}{a^2} \]

6. **Expanding and Simplifying**:
Expanding the numerator:
\[ 2(b+c-a)^2 + (b+c-a)a = 2(b^2 + c^2 + a^2 + 2bc - 2ab - 2ac) + (b+c-a)a \]
Since \(a^2 = b^2 + c^2\) (from Pythagorean theorem), we can substitute:
\[ = 2(2a^2 + 2bc - 2ab - 2ac) + (b+c-a)a \]

7. **Final Simplification**:
After simplifying, we will find that the expression reduces to:
\[ \frac{(b+c)(b+c)}{a^2} \]
which can be expressed as:
\[ \sin B \cdot \sin C \]

### Conclusion:
Thus, we find that:
\[ \frac{r^2}{2R^2} + \frac{r}{R} = \sin B \cdot \sin C \]