If angle of projection are `((pi)/4 + theta)` and `((pi)/4 -theta)` where `theta lt(pi)/4`, then the ration of horizontal ranges described by the projectile is (speed is same)-

Options

A. `2:1`
B. `1:2`
C. `1:1`
D. `2:3`

Explanation

To solve the problem of finding the ratio of horizontal ranges described by projectiles launched at angles \((\frac{\pi}{4} + \theta)\) and \((\frac{\pi}{4} - \theta)\), we can follow these steps:

### Step 1: Understand the Range Formula
The horizontal range \(R\) of a projectile launched with an initial speed \(v\) at an angle \(\theta\) is given by the formula:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
where \(g\) is the acceleration due to gravity.

### Step 2: Calculate the Range for Each Angle
1. **For the angle \((\frac{\pi}{4} + \theta)\):**
\[ R_1 = \frac{v^2 \sin(2(\frac{\pi}{4} + \theta))}{g} = \frac{v^2 \sin(\frac{\pi}{2} + 2\theta)}{g} \]
Using the identity \(\sin(\frac{\pi}{2} + x) = \cos(x)\):
\[ R_1 = \frac{v^2 \cos(2\theta)}{g} \]

2. **For the angle \((\frac{\pi}{4} - \theta)\):**
\[ R_2 = \frac{v^2 \sin(2(\frac{\pi}{4} - \theta))}{g} = \frac{v^2 \sin(\frac{\pi}{2} - 2\theta)}{g} \]
Using the identity \(\sin(\frac{\pi}{2} - x) = \cos(x)\):
\[ R_2 = \frac{v^2 \cos(2\theta)}{g} \]

### Step 3: Find the Ratio of the Ranges
Now we can find the ratio of the two ranges:
\[ \frac{R_1}{R_2} = \frac{\frac{v^2 \cos(2\theta)}{g}}{\frac{v^2 \sin(2\theta)}{g}} = \frac{\cos(2\theta)}{\sin(2\theta)} = \cot(2\theta) \]

### Step 4: Simplify the Expression
Since \(\theta < \frac{\pi}{4}\), we can conclude that:
\[ \cot(2\theta) = \frac{1}{\tan(2\theta)} \]

### Conclusion
The ratio of the horizontal ranges described by the projectiles is:
\[ \frac{R_1}{R_2} = \cot(2\theta) \]