if the tangent to the curve `x=a(theta+sintheta) , y=a(1+costheta)` at `theta=pi/3` makes an angle `alpha` x=axis then `alpha`

Options

A. `(pi)/(3)`
B. `(2pi)/(3)`
C. `(pi)/(6)`
D. `(5pi)/(6)`

Answer

To solve the problem, we need to find the angle \(\alpha\) that the tangent to the curve makes with the x-axis at \(\theta = \frac{\pi}{3}\). The curve is given by the parametric equations:

\[ x = a(\theta + \sin \theta) \]
\[ y = a(1 + \cos \theta) \]

### Step 1: Differentiate \(x\) and \(y\) with respect to \(\theta\)

We need to find \(\frac{dy}{dx}\), which requires us to differentiate both \(x\) and \(y\) with respect to \(\theta\).

1. Differentiate \(x\):
\[ \frac{dx}{d\theta} = a(1 + \cos \theta) \]

2. Differentiate \(y\):
\[ \frac{dy}{d\theta} = a(-\sin \theta) \]

### Step 2: Find \(\frac{dy}{dx}\)

Using the chain rule, we can find \(\frac{dy}{dx}\) as follows:

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(-\sin \theta)}{a(1 + \cos \theta)} \]

The \(a\) cancels out:

\[ \frac{dy}{dx} = \frac{-\sin \theta}{1 + \cos \theta} \]

### Step 3: Evaluate \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{3}\)

Now we substitute \(\theta = \frac{\pi}{3}\):

1. Calculate \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\)
2. Calculate \(\cos \frac{\pi}{3} = \frac{1}{2}\)

Substituting these values into \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{-\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} = \frac{-\frac{\sqrt{3}}{2}}{\frac{3}{2}} = -\frac{\sqrt{3}}{3} \]

### Step 4: Find the angle \(\alpha\)

The slope of the tangent line is given by:

\[ \tan \alpha = \frac{dy}{dx} = -\frac{\sqrt{3}}{3} \]

To find \(\alpha\), we take the arctangent:

\[ \alpha = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) \]

The angle whose tangent is \(-\frac{\sqrt{3}}{3}\) is \(-\frac{\pi}{6}\) or \(\frac{5\pi}{6}\) (in the second quadrant).

Thus, we have:

\[ \alpha = \frac{5\pi}{6} \]

### Final Answer

The angle \(\alpha\) that the tangent to the curve makes with the x-axis at \(\theta = \frac{\pi}{3}\) is:

\[ \alpha = \frac{5\pi}{6} \]