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 $k$?
 A: Writing $x = 2y$, we obtain
$$\begin{align}
\frac{1-\cos x + k\sin x}{\sin x + k(1+\cos x)} &= \frac{(1-\cos (2y)) + k\sin (2y)}{\sin(2y) + k(1+\cos(2y))}\\
&= \frac{2\sin^2 y + 2k\sin y\cos y}{2\sin y\cos y + 2k\cos^2 y}\\
&= \frac{\sin y}{\cos y}\cdot\frac{\sin y+k\cos y}{\sin y + k\cos y}\\
&= \tan y
\end{align}$$
using the double-angle formulae $\sin (2y) = 2\sin y\cos y$ and $\cos (2y) = \cos^2 y - \sin^2 y$.
A: If $\frac{A}{B}=\frac{C}{D}$, for example $\frac{2}{3}=\frac{4}{6}$, then both fractions are also equal to $\frac{A+C}{B+D}$, or $\frac{6}{9}$ in my example.
Since $\frac{C}{D}=\frac{kC}{kD}$, it also equals $\frac{A+kC}{B+kD}$.
All you have to do is check that $A/B=C/D$.
A: Transpose one term and split 
$$  \sin^2 x + \cos^2 x = 1 $$
into two terms and place them in numerator (N) and denominator (D) to tally by cross multiplication
$$ \dfrac {1- \cos x }{\sin x } = \dfrac {\sin x }{1 + \cos x }=  \dfrac {k \sin x }{k(1 + \cos x )} = \dfrac {1- \cos x + k \sin x }{\sin x + k(1 + \cos x )}=\tan \dfrac{x}2 $$
where we have multiplied  both N and D of central fraction by a common factor $k$ and added up separately the N and D (which is a valid procedure with fractions)  leaving its value unchanged for all values of $k$.
To learn further about this interesting property, I am tempted to say, it can also be seen equal to $ k^{th}$ root of:
$$  \dfrac {(1- \cos x)^k + \sin ^{k} x }{ \sin ^k x  + { (1 + \cos x) }^k } = \tan \dfrac{x}2$$ for all values of $k$, a result burnt into my memory from Hall & Knight's Algebra text-book.
EDIT1:
As each fraction equals $\tan \dfrac{x}2$ it is included to show independence of $k$ directly.
A: 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.
