Prove that the tangent to the curve x=a(1- cos t), y=a(1

Prove that the tangent to the curve $x=a(1-\cos t),\ y=a(1-\sin t)$ at an arbitrary point $M$ forms with the coordinate axes a triangle of constant perimeter.

Step 1: Slope and tangent line

\[ \dot x=a\sin t,\quad \dot y=-a\cos t,\qquad k=\frac{dy}{dx}=-\frac{\cos t}{\sin t}. \tag{1} \]

Substituting into $y-a(1-\sin t)=k[x-a(1-\cos t)]$, multiplying by $\sin t$, and simplifying the right side with $\cos^2t+\sin^2t=1$: \[\cos t\cdot x+\sin t\cdot y = a(\cos t+\sin t-1). \tag{2}\]

Step 2: Axis intercepts, via the half-angle $\tau=t/2$

With $1-\cos t=2\sin^2\tau$, $1-\sin t=(\cos\tau-\sin\tau)^2$, the numerator of both intercepts factors as $\cos t+\sin t-1=2\sin\tau(\cos\tau-\sin\tau)$ (using $\cos t=1-2\sin^2\tau$, $\sin t=2\sin\tau\cos\tau$); dividing by $\cos t=(\cos\tau-\sin\tau)(\cos\tau+\sin\tau)$ and by $\sin t=2\sin\tau\cos\tau$ respectively:

\[ x_{A} = \frac{2a\sin\tau}{\cos\tau+\sin\tau}, \qquad y_{B} = \frac{a(\cos\tau-\sin\tau)}{\cos\tau}. \tag{3} \]

Step 3: The sum $x_A+y_B$

Over the common denominator $\cos\tau(\cos\tau+\sin\tau)$, the numerator $2\sin\tau\cos\tau+(\cos\tau-\sin\tau)(\cos\tau+\sin\tau)=2\sin\tau\cos\tau+\cos^2\tau-\sin^2\tau=\sin2\tau+\cos2\tau=\sin t+\cos t$ (using $2\tau=t$), and the denominator $\cos\tau(\cos\tau+\sin\tau)=\cos^2\tau+\sin\tau\cos\tau=\dfrac{1+\cos t+\sin t}{2}$, so

\[ x_{A}+y_{B} = \frac{a(\cos t+\sin t)}{\dfrac{1+\cos t+\sin t}{2}} = \frac{2a(\cos t+\sin t)}{1+\cos t+\sin t}. \tag{4} \]

Step 4: The hypotenuse $AB$

Since $(\cos\tau-\sin\tau)^2(\cos\tau+\sin\tau)^2=(\cos^2\tau-\sin^2\tau)^2=\cos^2t$ and $4\sin^2\tau\cos^2\tau=\sin^2t$, \[x_A^2+y_B^2 = \frac{a^2(\sin^2t+\cos^2t)}{\cos^2\tau(\cos\tau+\sin\tau)^2} = \frac{a^2}{\cos^2\tau(\cos\tau+\sin\tau)^2}, \qquad |AB| = \frac{a}{\cos\tau(\cos\tau+\sin\tau)} = \frac{2a}{1+\cos t+\sin t}. \tag{5}\]

Step 5: The perimeter

\[ P = x_A+y_B+|AB| = \frac{2a(\cos t+\sin t)}{1+\cos t+\sin t}+\frac{2a}{1+\cos t+\sin t} = \frac{2a(1+\cos t+\sin t)}{1+\cos t+\sin t} = 2a. \tag{6} \]

Step 6: Verification

\[t=0.3,\,1.0,\,2.0,\,-0.5,\ a=1\text{: numerically computed } x_A+y_B+|AB| = 2.00000 \text{ in every case}\ \checkmark\]