Question:

Step 1: Identify the condition for the point on the curve. We need to find the point where the $x$-coordinate equals the $y$-coordinate, which means: $$\cos t + \ln\left(\tan\frac{t}{2}\right) = \sin t$$ Step 2: Verify that a solution exists. While finding the exact value of $t$ analytically is difficult, we proceed by computing the derivatives needed for the tangent length formula, which will turn out to be independent of the specific value of $t$ (as long as $\sin t > 0$). Step 3: Compute the derivative $\frac{dy}{dt}$. From the parametric equation $y = \sin t$: $$\frac{dy}{dt} = \cos t$$ Step 4: Compute the derivative $\frac{dx}{dt}$. From the parametric equation $x = \cos t + \ln\left(\tan\frac{t}{2}\right)$, we differentiate term by term: $$\frac{dx}{dt} = -\sin t + \frac{d}{dt}\left[\ln\left(\tan\frac{t}{2}\right)\right]$$ Using the chain rule: $$\frac{d}{dt}\left[\ln\left(\tan\frac{t}{2}\right)\right] = \frac{1}{\tan(t/2)} \cdot \sec^2\left(\frac{t}{2}\right) \cdot \frac{1}{2} = \frac{1}{2\sin(t/2)\cos(t/2)}$$ Using the double angle formula $\sin t = 2\sin(t/2)\cos(t/2)$: $$\frac{dx}{dt} = -\sin t + \frac{1}{\sin t}$$ Combining over a common denominator: $$\frac{dx}{dt} = \frac{-\sin^2 t + 1}{\sin t} = \frac{\cos^2 t}{\sin t}$$ Step 5: Compute the slope $\frac{dy}{dx}$. Using the chain rule for parametric equations: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t}{\cos^2 t/\sin t} = \frac{\cos t \cdot \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} = \tan t$$ Step 6: Apply the tangent length formula. The length of the tangent from a point $(x, y)$ on the curve to the $x$-axis is given by: $$L = y\sqrt{1 + \left(\frac{dx}{dy}\right)^2}$$ Since $\frac{dy}{dx} = \tan t$, we have $\frac{dx}{dy} = \cot t$, so: $$L = y\sqrt{1 + \cot^2 t}$$ Using the trigonometric identity $1 + \cot^2 t = \csc^2 t = \frac{1}{\sin^2 t}$: $$L = \sin t \cdot \sqrt{\frac{1}{\sin^2 t}} = \sin t \cdot \frac{1}{|\sin t|}$$ For $\sin t > 0$ (which holds at the point where $x = y$): $$L = \sin t \cdot \frac{1}{\sin t} = 1$$ Step 7: State the final answer. The length of the tangent drawn to the curve at the point where the $x$-coordinate equals the $y$-coordinate is $\boxed{1}$. The answer is **Option 1: 1**.