Find dy/dx, if x=log⁡(tan⁡t) and y=log⁡(sin⁡t)

The correct option is cos²t.

To find the derivative of $\frac{dy}{dx}$ for the given parametric equations, we use the chain rule for parametric differentiation:
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

Step 1: Find the derivative of $x$ with respect to $t$ ($dx/dt$)
We are given:
$$x=\log (\tan t)$$
Applying the chain rule:
$$\frac{dx}{dt}=\frac{1}{\tan t}\cdot \frac{d}{dt}(\tan t)$$
Since $\frac{d}{dt}(\tan t)={\sec }^{2}t$, we have:
$$\frac{dx}{dt}=\frac{1}{\tan t}\cdot {\sec }^{2}t$$
Let us simplify this trigonometric expression:
$$\frac{dx}{dt}=\frac{\cos t}{\sin t}\cdot \frac{1}{{\cos }^{2}t}=\frac{1}{\sin t\cos t}$$

Step 2: Find the derivative of $y$ with respect to $t$ ($dy/dt$)
We are given:
$$y=\log (\sin t)$$
Applying the chain rule:
$$\frac{dy}{dt}=\frac{1}{\sin t}\cdot \frac{d}{dt}(\sin t)$$
Since $\frac{d}{dt}(\sin t)=\cos t$, we have:
$$\frac{dy}{dt}=\frac{\cos t}{\sin t}$$

Step 3: Calculate $\frac{dy}{dx}$
Substitute the values from Step 1 and Step 2 into the parametric derivative formula:
$$\frac{dy}{dx}=\frac{\frac{\cos t}{\sin t}}{\frac{1}{\sin t\cos t}}$$
Simplify the division of fractions by multiplying by the reciprocal:
$$\frac{dy}{dx}=\frac{\cos t}{\sin t}\cdot (\sin t\cos t)$$
Canceling the common term $\sin t$ in the numerator and denominator:
$$\frac{dy}{dx}=\cos t\cdot \cos t={\cos }^{2}t$$