Show that the length of the tangent of the curve x=a (ln

Show that the length of the tangent of the curve $x=a\!\left(\ln\tan\dfrac{t}{2}+\cos t\right),\ y=a\sin t$ is constant.

Step 1: The tangent-length formula

For a parametric curve, $T=y\sqrt{1+(dx/dy)^2}$ with $dx/dy=\dot x/\dot y$.

Step 2: Compute $\dot y$

\[ \dot y = a\cos t. \tag{1} \]

Step 3: Compute $\dot x$

Differentiating the two terms separately: \[\frac{d}{dt}\ln\tan\frac t2 = \frac{1}{\tan(t/2)}\cdot\sec^2\frac t2\cdot\frac12 = \frac{\cos(t/2)}{2\sin(t/2)\cos^2(t/2)} = \frac{1}{2\sin(t/2)\cos(t/2)} = \frac{1}{\sin t}, \qquad \frac{d}{dt}\cos t = -\sin t. \tag{2}\]

\[ \dot x = a\!\left(\frac1{\sin t}-\sin t\right) = a\cdot\frac{1-\sin^2t}{\sin t} = a\cdot\frac{\cos^2t}{\sin t}. \tag{3} \]

Step 4: The ratio $dx/dy$

\[ \frac{dx}{dy} = \frac{\dot x}{\dot y} = \frac{a\cos^2t/\sin t}{a\cos t} = \frac{\cos t}{\sin t} = \cot t. \tag{4} \]

Step 5: Compute the tangent length

For $t\in(0,\pi)$, so that $y=a\sin t>0$, and using $1+\cot^2t=\csc^2t$: \[1+\left(\frac{dx}{dy}\right)^2 = \csc^2t = \frac{1}{\sin^2t} \;\Longrightarrow\; T = a\sin t\cdot\frac{1}{\sin t} = a. \tag{5}\]

Step 6: Verification

\[a=3,\ t=1.0\text{: } dx/dy=\cot(1.0)\approx0.64209\ (\text{matches the raw parametric ratio }\dot x/\dot y\text{ exactly}),\ T=3.00000=a\ \checkmark\]