6. If \( x=a(\cos t+\log \tan t 2), y=a \sin t \) find: \( d y / d x \)

To find \(\frac{dy}{dx}\) for the given parametric equations: \[ x = a\left(\cos t + \log(2 \tan t)\right), \quad y = a \sin t, \] we'll use the chain rule for derivatives in parametric form: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}. \] ### Step 1: Compute \(\frac{dy}{dt}\) Given: \[ y = a \sin t, \] the derivative with respect to \(t\) is: \[ \frac{dy}{dt} = a \cos t. \] ### Step 2: Compute \(\frac{dx}{dt}\) Given: \[ x = a\left(\cos t + \log(2 \tan t)\right), \] the derivative with respect to \(t\) is: \[ \frac{dx}{dt} = a\left(-\sin t + \frac{d}{dt} \log(2 \tan t)\right). \] To find \(\frac{d}{dt} \log(2 \tan t)\), use the chain rule: \[ \frac{d}{dt} \log(2 \tan t) = \frac{1}{2 \tan t} \cdot 2 \sec^2 t = \frac{\sec^2 t}{\tan t}. \] Simplify: \[ \frac{\sec^2 t}{\tan t} = \frac{1}{\cos^2 t} \cdot \frac{\cos t}{\sin t} = \frac{1}{\sin t \cos t}. \] Thus: \[ \frac{dx}{dt} = a\left(-\sin t + \frac{1}{\sin t \cos t}\right). \] ### Step 3: Compute \(\frac{dy}{dx}\) Now, substitute the derivatives into the chain rule: \[ \frac{dy}{dx} = \frac{a \cos t}{a\left(-\sin t + \frac{1}{\sin t \cos t}\right)} = \frac{\cos t}{-\sin t + \frac{1}{\sin t \cos t}}. \] To simplify, find a common denominator for the terms in the denominator: \[ -\sin t + \frac{1}{\sin t \cos t} = \frac{-\sin^2 t \cos t + 1}{\sin t \cos t}. \] Thus: \[ \frac{dy}{dx} = \frac{\cos t}{\frac{-\sin^2 t \cos t + 1}{\sin t \cos t}} = \frac{\cos^2 t \sin t}{1 - \sin^2 t \cos t}. \] ### Final Answer \[ \boxed{\dfrac{dy}{dx} = \dfrac{\cos t}{\, -\,\sin t\,+\,\dfrac{1}{\,\cos t\,\sin t\,}}} \] Alternatively, it can be written as: \[ \frac{dy}{dx} = \frac{\sin t \cos^2 t}{1 - \sin^2 t \cos t} \]