Geometric construction showing independency of constant By means of a geometric construction show that 

$$\frac{1-\cos t + k\sin t}{\sin t+ k \ (1+\cos t)} $$

does not depend on $k$
To be honest, I don't think this question is of very much meaning.

Note that $\tan(\frac{x}{2})=\frac{1-\cos t}{\sin t}=\frac{k \sin t}{k(1+\cos t)}=\frac{1-\cos t+k \sin t}{\sin t+k(1-\cos t)}$.

So it suffices to make a geometric construction to show $tan\left(\frac{x}{2}\right)=\frac{1-\cos t}{\sin t}$.

The following image shows the construction.
$$\frac{1-\cos x + k\sin x}{\sin x + k(1+\cos x)}=\\
\frac{2\sin^2(\frac x2) + k2\sin (\frac x2)\cos (\frac x2)}{2\sin (\frac x2)\cos (\frac x2) + 2k\cos^2(\frac x2)}=\\
\frac22.\frac{\sin (\frac x2)}{\cos (\frac x2)}.\frac{\sin(\frac x2) + k\cos (\frac x2)}{\sin (\frac x2) + k\cos(\frac x2)}=1.\tan(\frac{x}{2}).1=\\\tan(\frac{x}{2})$$