The length of tangent to the curve `x=a(cost+log tan.(t)/(2)),y=a(sint),` is

Options

A. ax
B. ay
C. a
D. xy

Answer

To find the length of the tangent to the curve given by the parametric equations \( x = a \left( \cos t + \log \tan \frac{t}{2} \right) \) and \( y = a \sin t \), we will follow these steps:

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

1. Differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = a \cos t \]

2. Differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = a \left( -\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2} \sec^2 \frac{t}{2} \cdot \frac{1}{2} \right) \]
Here, we apply the chain rule for the logarithmic differentiation of \( \log \tan \frac{t}{2} \).

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

Using the chain rule, we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{a \cos t}{a \left( -\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2} \sec^2 \frac{t}{2} \cdot \frac{1}{2} \right)} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\cos t}{-\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2} \sec^2 \frac{t}{2}} \]

### Step 3: Calculate the length of the tangent

The length of the tangent \( L \) can be calculated using the formula:
\[ L = y_1 \sqrt{1 + m^2} / m \]
Where \( y_1 = a \sin t \) and \( m = \frac{dy}{dx} \).

1. Substitute \( y_1 \) and \( m \):
\[ L = a \sin t \sqrt{1 + \left( \frac{\cos t}{-\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2} \sec^2 \frac{t}{2}} \right)^2} \cdot \frac{-\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2} \sec^2 \frac{t}{2}}{\cos t} \]

2. Simplify the expression:
\[ L = a \sin t \cdot \sqrt{\sec^2 t} \cdot \frac{-\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2} \sec^2 \frac{t}{2}}{\cos t} \]
Recognizing that \( \sqrt{1 + \tan^2 t} = \sec t \).

3. Final simplification leads to:
\[ L = a \]

### Conclusion

Thus, the length of the tangent to the curve is \( a \).

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