Question

\(x \sin y+y \sin x=\pi\) \(\sin y+x \cos y \frac{d y}{d x}+\frac{d y}{d x} \sin x+y \cos x=0\) \(\frac{d y}{d x}=\frac{-(\sin y+y \cos x)}{x \cos y+\sin x}=-1\) eqn of tangent \(\rightarrow y=-x+\pi\)

Short answer

Answer: The equation of the tangent to the curve is \(y=-x+\pi\).

Step-by-step solution

Equation 1

\(a \sin b+b \sin a=\pi\) Now we need to find the value of (a, b) by solving Equation 1. #Step 2: Applying the first order derivative to the point (a, b)# We know that the slope of the tangent at the point (a, b) is -1. The first-order derivative represents the slope of the tangent at any point on the curve, and since the tangent touches the curve at (a, b), we can write the given first-order derivative as:

Equation 2

\(\frac{-(\sin b+b \cos a)}{a \cos b+\sin a}=-1\) Now, we will solve Equations 1 and 2 simultaneously to get the values of a and b. #Step 3: Solving Equations 1 and 2 simultaneously# We already have Equation 1:

Equation 1 (repeated)

\(a \sin b+b \sin a=\pi\) From Equation 2, we can write:

Equation 2 modified

\((\sin b+b \cos a)=-(a \cos b+\sin a)\) Add both Equation 1 and modified Equation 2:

Combined Equation

\((a \sin b+b \sin a) + (\sin b+b \cos a) = \pi\) #Step 4: Calculate point (a, b)# Upon solving Combined Equation:

Point (a, b) calculation

\(a \sin b+b \sin a + \sin b+b \cos a = \pi \Rightarrow a=-b+\pi\) Now we can plug a into the tangent equation to find the final equation of the tangent: #Step 5: Calculate the equation of the tangent# Given that the slope of the tangent is -1, we can write the tangent equation as:

Tangent Equation

\(y=-x+c\) Substitute (a, b) into the tangent equation:

Tangent Equation with (a, b)

\(b=-a+c\) Since \(a=-b+\pi\), we can write the tangent equation as:

Final Tangent Equation

\(y=-x+\pi\) Thus, the equation of the tangent to the curve represented by the given equation is \(y=-x+\pi\).