If ` x=2 log cot t and y= tan t + cot t, ` then `( dy)/(dx) sin 2 t +1=`

Options

A. ` cos^2 t `
B. ` sin^2 t `
C. ` cos 2 t`
D. ` 2 cos^2 t `

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) given the parametric equations \(x = 2 \log(\cot t)\) and \(y = \tan t + \cot t\), and then evaluate \(\frac{dy}{dx} \sin(2t) + 1\).

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

Given:
\[ y = \tan t + \cot t \]

We differentiate \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = \frac{d}{dt}(\tan t) + \frac{d}{dt}(\cot t) \]
Using the derivatives \(\frac{d}{dt}(\tan t) = \sec^2 t\) and \(\frac{d}{dt}(\cot t) = -\csc^2 t\), we have:
\[ \frac{dy}{dt} = \sec^2 t - \csc^2 t \]

### Step 2: Differentiate \(x\) with respect to \(t\)

Given:
\[ x = 2 \log(\cot t) \]

We differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = 2 \cdot \frac{d}{dt}(\log(\cot t)) \]
Using the chain rule, we know \(\frac{d}{dt}(\log(u)) = \frac{1}{u} \cdot \frac{du}{dt}\):
\[ \frac{dx}{dt} = 2 \cdot \frac{1}{\cot t} \cdot \frac{d}{dt}(\cot t) \]
Now, \(\frac{d}{dt}(\cot t) = -\csc^2 t\), so:
\[ \frac{dx}{dt} = 2 \cdot \frac{-\csc^2 t}{\cot t} = -2 \frac{\csc^2 t}{\cot t} \]

### Step 3: Find \(\frac{dy}{dx}\)

Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\sec^2 t - \csc^2 t}{-2 \frac{\csc^2 t}{\cot t}} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\sec^2 t - \csc^2 t}{-2 \frac{\csc^2 t}{\cot t}} = \frac{\cot t (\sec^2 t - \csc^2 t)}{-2 \csc^2 t} \]

### Step 4: Simplify \(\frac{dy}{dx}\)

We can express \(\sec^2 t\) and \(\csc^2 t\) in terms of sine and cosine:
\[ \sec^2 t = \frac{1}{\cos^2 t}, \quad \csc^2 t = \frac{1}{\sin^2 t} \]
Thus:
\[ \frac{dy}{dx} = \frac{\cot t \left(\frac{1}{\cos^2 t} - \frac{1}{\sin^2 t}\right)}{-2 \cdot \frac{1}{\sin^2 t}} = \frac{\cot t \left(\frac{\sin^2 t - \cos^2 t}{\sin^2 t \cos^2 t}\right)}{-2 \cdot \frac{1}{\sin^2 t}} = \frac{\cot t (\sin^2 t - \cos^2 t)}{-2 \cos^2 t} \]

### Step 5: Multiply by \(\sin(2t)\) and add 1

Now we need to compute:
\[ \frac{dy}{dx} \sin(2t) + 1 \]
Using the identity \(\sin(2t) = 2 \sin t \cos t\):
\[ \frac{dy}{dx} \sin(2t) = \frac{\cot t (\sin^2 t - \cos^2 t)}{-2 \cos^2 t} \cdot 2 \sin t \cos t \]
This simplifies to:
\[ = \frac{\cot t (\sin^2 t - \cos^2 t) \sin t}{-\cos^2 t} \]
Now, substituting \(\cot t = \frac{\cos t}{\sin t}\):
\[ = \frac{\frac{\cos t}{\sin t} (\sin^2 t - \cos^2 t) \sin t}{-\cos^2 t} = \frac{\cos t (\sin^2 t - \cos^2 t)}{-\cos^2 t} = -\frac{\sin^2 t - \cos^2 t}{\cos t} \]

Adding 1:
\[ -\frac{\sin^2 t - \cos^2 t}{\cos t} + 1 \]

### Final Result

The final expression simplifies to:
\[ 1 + \cos(2t) = 2 \cos^2 t \]