If `tantheta=(x-y)/(x+y)`,then value of `sintheta` is eqaul to (If `0^(@)lethetale90^(@))`

Options

A. `(x-y)/(sqrt(2(x^(2)+y^(2))))`
B. `(x+y)/(sqrt(2(x^(2)+y^(2))))`
C. `(x+y)/(sqrt(2(x^(2)-y^(2))))`
D. `(x-y)/(sqrt(2(x^(2)-y^(2))))`

Answer

To find the value of \(\sin \theta\) given that \(\tan \theta = \frac{x - y}{x + y}\), we can follow these steps:

### Step 1: Understand the relationship between sine, cosine, and tangent
We know that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
From the given equation, we can express \(\sin \theta\) in terms of \(\tan \theta\) and \(\cos \theta\).

### Step 2: Set up the right triangle
We can visualize \(\tan \theta\) as the ratio of the opposite side (perpendicular) to the adjacent side (base) of a right triangle:
- Let the opposite side (perpendicular) be \(x - y\).
- Let the adjacent side (base) be \(x + y\).

### Step 3: Calculate the hypotenuse
Using the Pythagorean theorem, the hypotenuse \(h\) can be calculated as:
\[ h = \sqrt{(x - y)^2 + (x + y)^2} \]
Now, we will expand this expression:
\[ h = \sqrt{(x - y)^2 + (x + y)^2} = \sqrt{(x^2 - 2xy + y^2) + (x^2 + 2xy + y^2)} = \sqrt{2x^2 + 2y^2} \]
This simplifies to:
\[ h = \sqrt{2(x^2 + y^2)} \]

### Step 4: Calculate \(\sin \theta\)
Now, we can find \(\sin \theta\) using the definition:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x - y}{\sqrt{2(x^2 + y^2)}} \]

### Final Result
Thus, the value of \(\sin \theta\) is:
\[ \sin \theta = \frac{x - y}{\sqrt{2(x^2 + y^2)}} \]