The equation of the line tangent to the curve x siny + ysinx = `pi` at the point `((pi)/(2),(pi)/(2))` is

Options

A. `3x+y=2pi`
B. `x-y=0`
C. `2x-y=pi//2`
D. `x+y=pi`

Answer

To find the equation of the line tangent to the curve \( x \sin y + y \sin x = \pi \) at the point \( \left( \frac{\pi}{2}, \frac{\pi}{2} \right) \), we will follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the equation of the curve:
\[ x \sin y + y \sin x = \pi \]
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x \sin y) + \frac{d}{dx}(y \sin x) = \frac{d}{dx}(\pi) \]
Using the product rule and chain rule, we get:
\[ \sin y + x \cos y \frac{dy}{dx} + \sin x \frac{dy}{dx} + y \cos x = 0 \]

### Step 2: Substitute the point into the derivative
Now we substitute the point \( \left( \frac{\pi}{2}, \frac{\pi}{2} \right) \) into the differentiated equation:
\[ \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) \frac{dy}{dx} + \sin\left(\frac{\pi}{2}\right) \frac{dy}{dx} + \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) = 0 \]
This simplifies to:
\[ 1 + 0 \cdot \frac{dy}{dx} + \frac{dy}{dx} + 0 = 0 \]
Thus, we have:
\[ 1 + \frac{dy}{dx} = 0 \]
This leads to:
\[ \frac{dy}{dx} = -1 \]

### Step 3: Use the point-slope form to find the equation of the tangent line
Now that we have the slope of the tangent line, which is \( -1 \), we can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = -1 \), \( x_1 = \frac{\pi}{2} \), and \( y_1 = \frac{\pi}{2} \):
\[ y - \frac{\pi}{2} = -1\left(x - \frac{\pi}{2}\right) \]
This simplifies to:
\[ y - \frac{\pi}{2} = -x + \frac{\pi}{2} \]
Rearranging gives:
\[ x + y = \pi \]

### Final Answer
The equation of the tangent line is:
\[ \boxed{x + y = \pi} \]