Given that `theta` is acute and then `sin theta=(3)/(5)`. Let x, y be positive real number such that `3(x-y)=1`, then one set of solutions for x and y expressed in terms of `theta` is given by :

Options

A. A)`x = sec theta, y =" cosec "theta`
B. B)`x=cot theta, y= tan theta`
C. C)`x="cosec "theta, y=cot theta`
D. D)`x= sec theta, y=tan theta`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the given information
We are given that \( \sin \theta = \frac{3}{5} \) and that \( \theta \) is an acute angle. We also have the equation \( 3(x - y) = 1 \).

### Step 2: Draw a right triangle
To visualize the problem, we can draw a right triangle where:
- The angle \( \theta \) is one of the acute angles.
- The side opposite to \( \theta \) (perpendicular) is 3.
- The hypotenuse is 5.

### Step 3: Find the length of the base
Using the Pythagorean theorem, we can find the length of the base (adjacent side):
\[ \text{Base}^2 + \text{Perpendicular}^2 = \text{Hypotenuse}^2 \]
Let the base be \( b \):
\[ b^2 + 3^2 = 5^2 \\ b^2 + 9 = 25 \\ b^2 = 16 \\ b = 4 \]
So, the base (adjacent side) is 4.

### Step 4: Assign values to \( x \) and \( y \)
We need to express \( x \) and \( y \) in terms of trigonometric functions. Since we have the values of the sides, we can use:
- \( x = \csc \theta \) (which is \( \frac{1}{\sin \theta} = \frac{5}{3} \))
- \( y = \cot \theta \) (which is \( \frac{\text{base}}{\text{perpendicular}} = \frac{4}{3} \))

### Step 5: Substitute \( x \) and \( y \) into the equation
Now we substitute \( x \) and \( y \) into the equation \( 3(x - y) = 1 \):
\[ 3\left(\csc \theta - \cot \theta\right) = 1 \]
Substituting the values:
\[ 3\left(\frac{5}{3} - \frac{4}{3}\right) = 1 \\ 3\left(\frac{1}{3}\right) = 1 \\ 1 = 1 \]
This confirms that our values for \( x \) and \( y \) satisfy the equation.

### Conclusion
Thus, one set of solutions for \( x \) and \( y \) expressed in terms of \( \theta \) is:
\[ x = \csc \theta \quad \text{and} \quad y = \cot \theta \]