Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ I would like to try and prove
$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$$
using $LHS=RHS$ methods, i.e. pick a side and rewrite it to make it identical to the other side.
I found a quick way by doing this:
$$LHS = \frac{1+\sin x}{\cos x} = \frac{1+\sin x}{\cos x} \cdot \frac{1 - \tan x + \sec x}{1 - \tan x + \sec x}= \frac{1+\sin x+\cos x}{1-\sin x+\cos x} = RHS$$
but I feel that this is not a good way because I am manipulating the denominator of the LHS somewhat artificially, because I know it must be, in the end, $1-\sin x+\cos x$.
Does anyone have a better way of doing this?
 A: $\frac{1+\sin x}{\cos x}$
$=\frac{(1+\sin x)(1-\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{1-\sin^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{\cos^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{(\cos x)(1+\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}$
A: Using $\cos^2x+\sin^2x=1$, we have 
$$ (1+\sin x)(1-\sin x+\cos x)=1-\sin^2x+(1+\sin x)\cos x=\cos x(\cos x+1+\sin x).$$
Now divide the LHS by $1-\sin x+\cos x$ and the RHS by $\cos x$ to get the result.
A: We'll work on the RHS. We'll multiply $\frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ with the conjugate of the denominator. We'll do this from right to left, i.e. we pick the RHS and walk through to LHS.
Solution 1:
$$\require{cancel}\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1+\sin x-\cos x}{1+\sin x-\cos x}\\&=\frac{1+\sin x\cancel{-\cos x}+\sin x+\sin^{2}x\cancel{-\sin{x}\cos{x}}\cancel{+\cos x}\cancel{+\sin{x}\cos{x}}-\cos^{2}x}{1\cancel{+\sin x}\cancel{-\cos x}\cancel{-\sin x}-\sin^{2}x+\sin{x}\cos{x}\cancel{+\cos x}+\sin{x}\cos{x}-\cos^{2}x}\\&=\frac{1+2\sin x+\sin^{2}x-\cos^{2}x}{1-\sin^{2}x+2\sin{x}\cos{x}-\cos^{2}x}\\&=\frac{1+2\sin x+\sin^{2}x-\cos^{2}x}{\cancel{\cos^{2}x}+2\sin{x}\cos{x}\cancel{-\cos^{2}x}}\\&=\frac{1+2\sin x+\sin^{2}x-\cos^{2}x}{2\sin{x}\cos{x}}\\&=\frac{\sin^{2}x\cancel{+\cos^{2}x}+2\sin x+\sin^{2}x\cancel{-\cos^{2}x}}{2\sin{x}\cos{x}}\\&=\frac{\cancel2\sin^{2}x+\cancel2\sin x}{\cancel2\sin{x}\cos{x}}\\&=\frac{\sin^{2}x+\sin x}{\sin{x}\cos{x}}\\&=\frac{\cancel{\sin{x}}\left(\sin x+1\right)}{\cancel{\sin{x}}\cos{x}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
Solution 2:
$$\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1-\sin x-\cos x}{1-\sin x-\cos x}\\&=\frac{1\cancel{-\sin x}\cancel{-\cos x}\cancel{+\sin x}-\sin^{2}x-\sin{x}\cos{x}\cancel{+\cos x}-\sin{x}\cos{x}-\cos^{2}x}{1-\sin x\cancel{-\cos x}-\sin x+\sin^{2}x\cancel{+\sin{x}\cos{x}}\cancel{+\cos x}\cancel{-\sin{x}\cos{x}}-\cos^{2}x}\\&=\frac{1-\sin^{2}x-2\sin{x}\cos{x}-\cos^{2}x}{1-2\sin x+\sin^{2}x-\cos^{2}x}\\&=\frac{\cancel{\sin^{2}}x+\cancel{\cos^{2}x}\cancel{-\sin^{2}x}-2\sin{x}\cos{x}\cancel{-\cos^{2}x}}{\sin^{2}x-2\sin x+\sin^{2}x}\\&=\frac{-\cancel2\sin x\cos x}{\cancel2\sin^{2}x-\cancel2\sin x}\\&=\frac{-\sin{x}\cos{x}}{\sin^{2}x-\sin x}\\&=\frac{\sin{x}\cos{x}}{\sin x-\sin^{2}x}\\&=\frac{\cancel{\sin{x}}\cos{x}}{\cancel{\sin x}(1-\sin x)}\\&=\frac{\cos{x}}{1-\sin{x}}\cdot\frac{1+\sin{x}}{1+\sin{x}}\\&=\frac{\cos{x}(1+\sin{x})}{1-\sin^2{x}}\\&=\frac{\cancelto{1}{\cos{x}}(1+\sin{x})}{\cancelto{\cos{x}}{\cos^2{x}}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
Solution 3:
$$\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1-\sin x+\cos x}{1-\sin x+\cos x}\\&=\frac{1+\sin x\cancel{-\cos x}+\sin x+\sin^{2}x\cancel{-\sin{x}\cos{x}}\cancel{+\cos x}\cancel{+\sin{x}\cos{x}}-\cos^{2}x}{1+\cancel{\sin x}\cancel{-\cos x}\cancel{-\sin x}-\sin^{2}x+\sin{x}\cos{x}\cancel{+\cos x}+\sin{x}\cos{x}-\cos^{2}x}\\&=\frac{1+2\cos x-\sin^{2}x+\cos^{2}x}{1-2\sin x+2\cos x+\sin^{2}x-2\sin{x}\cos{x}+\cos^{2}x}\\&=\frac{\cancel{\sin^{2}x}+\cos^{2}x+2\cos x\cancel{-\sin^{2}x}+\cos^{2}x}{\sin^{2}x+\cos^{2}x-2\sin x+2\cos x+\sin^{2}x-2\sin{x}\cos{x}+\cos^{2}x}\\&=\frac{\cancel2\cos^{2}x+\cancel2\cos x}{\cancel2\sin^{2}x+\cancel2\cos^{2}x-\cancel2\sin x+\cancel2\cos x-\cancel2\sin{x}\cos{x}}\\&=\frac{\cos^{2}x+\cos x}{\sin^{2}x+\cos^{2}x-\sin x+\cos x-\sin{x}\cos{x}}\\&=\frac{\cos{x}\left(1+\cos x\right)}{1-\sin x-\sin{x}\cos{x}+\cos x}\\&=\frac{\cos{x}\left(1+\cos x\right)}{1-\sin x+\cos{x}\left(1-\sin x\right)}\\&=\frac{\cos{x}\cancel{\left(1+\cos x\right)}}{\cancel{\left(1+\cos x\right)}\left(1-\sin x\right)}\\&=\frac{\cos{x}}{1-\sin{x}}\cdot\frac{1+\sin{x}}{1+\sin{x}}\\&=\frac{\cos{x}(1+\sin{x})}{1-\sin^2{x}}\\&=\frac{\cancelto{1}{\cos{x}}(1+\sin{x})}{\cancelto{\cos{x}}{\cos^2{x}}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
Solution 4:
$$\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1+\sin x+\cos x}{1+\sin x+\cos x}\\&=\frac{1+\sin x+\cos x+\sin x+\sin^{2}x+\sin{x}\cos{x}+\cos x+\sin{x}\cos{x}+\cos^{2}x}{1\cancel{+\sin x}+\cos x\cancel{-\sin x}-\sin^{2}x\cancel{-\sin{x}\cos{x}}+\cos x\cancel{+\sin{x}\cos{x}}+\cos^{2}x}\\&=\frac{1+2\sin x+2\cos x+\sin^{2}x+2\sin{x}\cos{x}+\cos^{2}x}{1+2\cos x-\sin^{2}x+\cos^{2}x}\\&=\frac{\cancelto{1}{2}+\cancel2\sin x+\cancel2\cos x+\cancel2\sin{x}\cos{x}}{\cancel2\cos x+\cancel2\cos^{2}x}\\&=\frac{1+\sin x+\cos x+\sin{x}\cos{x}}{\cos x+\cos^{2}x}\\&=\frac{1+\sin x+\cos{x}\left(1+\sin x\right)}{\cos{x}\left(1+\cos x\right)}\\&=\frac{\cancel{\left(1+\cos x\right)}\left(1+\sin x\right)}{\cos{x}\cancel{\left(1+\cos x\right)}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
I hope this helps.
A: For $b\not=0$ we have:
$$\begin{align*}\frac{1+a}{b} &= \frac{1+a+b}{1-a+b}\qquad&\iff \\
(1+a)(1-a+b) &= b(1+a+b)\qquad&\iff\\
1-a^2+b(a+1) &= b^2 + b(a+1)\qquad&\iff\\
a^2+b^2&=1
\end{align*}$$
So your equation is an alternate way to characterize $\cos^2 x+\sin^2 x = 1$
A: Let's call $a=\dfrac{1+s}c$
Replacing $1$ by $c^2+s^2$ gives $\ a=\dfrac{c^2+s^2+s}c=c+s\dfrac{1+s}c=c+sa\iff a(1-s)=c$
So we have also $a=\dfrac c{1-s}$
Now we can use the identity $$\frac pq=\frac rs=x\implies \frac{p+r}{q+s}=\frac{qx+sx}{q+s}=x$$
Thus $a=\dfrac{(1+s)+c}{c+(1-s)}$
