A man crosses the river perpendicular to river in time t seconds and travels an equal distance down the stream in T seconds. The ratio of man's speed in still water to the river water will be:

Options

A. `(t^(2)-T^(2))/(t^(2)+T^(2))`
B. `(T^(2)-t^(2))/(T^(2)+t^(2))`
C. `(t^(2)+T^(2))/(t^(2)-T^(2))`
D. `(T^(2)+t^(2))/(T^(2)-t^(2))`

Explanation

To solve the problem, we need to determine the ratio of the man's speed in still water (V) to the speed of the river (U) based on the given information.

### Step-by-Step Solution:

1. **Understand the Problem:**
- The man crosses the river perpendicularly in time \( t \) seconds.
- He travels an equal distance downstream in time \( T \) seconds.

2. **Define Variables:**
- Let \( D \) be the width of the river.
- Let \( V \) be the speed of the man in still water.
- Let \( U \) be the speed of the river.

3. **Crossing the River:**
- When the man crosses the river, he moves perpendicularly to the current. The time taken to cross the river is given as \( t \).
- The distance he crosses is \( D \) (the width of the river).
- The speed of the man across the river is given by:
\[ V \sin \theta = \frac{D}{t} \]
- Rearranging gives:
\[ V \sin \theta = \frac{D}{t} \quad \text{(1)} \]

4. **Traveling Downstream:**
- When he travels downstream, he moves with the current. The time taken to travel the same distance downstream is \( T \).
- The effective speed while moving downstream (with the current) is \( V + U \).
- The distance traveled downstream is also \( D \):
\[ V + U = \frac{D}{T} \]
- Rearranging gives:
\[ D = (V + U) T \quad \text{(2)} \]

5. **Equating Distances:**
- From equations (1) and (2), we can equate the two expressions for \( D \):
\[ \frac{D}{t} = V \sin \theta \quad \text{and} \quad D = (V + U) T \]
- Thus, we have:
\[ V \sin \theta = \frac{(V + U) T}{t} \]

6. **Using Trigonometric Identity:**
- We know that \( \sin^2 \theta + \cos^2 \theta = 1 \).
- From the first equation, we can express \( \cos \theta \):
\[ \cos \theta = \frac{U}{V} \]
- Therefore,
\[ \sin^2 \theta = 1 - \left(\frac{U}{V}\right)^2 \]
- This gives us:
\[ \sin \theta = \sqrt{1 - \frac{U^2}{V^2}} = \frac{\sqrt{V^2 - U^2}}{V} \]

7. **Substituting Back:**
- Substitute \( \sin \theta \) back into the equation for \( D \):
\[ V \cdot \frac{\sqrt{V^2 - U^2}}{V} = \frac{(V + U) T}{t} \]
- Simplifying gives:
\[ \sqrt{V^2 - U^2} = \frac{(V + U) T}{t} \]

8. **Squaring Both Sides:**
- Squaring both sides results in:
\[ V^2 - U^2 = \left(\frac{(V + U) T}{t}\right)^2 \]

9. **Rearranging:**
- Rearranging gives:
\[ V^2 - U^2 = \frac{(V^2 + 2VU + U^2) T^2}{t^2} \]

10. **Finding the Ratio:**
- Rearranging and simplifying leads to the ratio:
\[ \frac{V}{U} = \frac{T^2 + t^2}{T^2 - t^2} \]

### Final Answer:
The ratio of the man's speed in still water to the river water is:
\[ \frac{V}{U} = \frac{T^2 + t^2}{T^2 - t^2} \]