Independence of $\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$ from $k$. How to one can show that the value of the following expression
$$\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$$ 
doesn't depend to values of $k$?

We have $\frac{1-\cos x + k \sin x}{\sin x + k(1+\cos x)}$ and what bothers me at first sight is that $k$ multiplies $\sin$ in numerator but $\cos$ in denominator. So, let's turn $\cos$ into $\sin$:
$$
\begin{align}\frac{1-\cos x + k \sin x}{\sin x + k(1+\cos x)} &= \frac{1-\cos x + k \sin x}{\sin x + k(1+\cos x)}\cdot \frac{1-\cos x}{1-\cos x}\\ 
&= \frac{(1-\cos x)(1-\cos x + k \sin x)}{(1-\cos x)\sin x + k(1-\cos^2 x)}\\ &= \frac{(1-\cos x)(1-\cos x + k \sin x)}{(1-\cos x)\sin x + k\sin^2 x}\\ 
&= \frac{(1-\cos x)(1-\cos x + k \sin x)}{\sin x(1-\cos x + k\sin x)}\\
&= \frac{1-\cos x}{\sin x}
\end{align}
$$
EDIT: Exactly the same technique can be used to solve Narasimham's problem as well.
