# Thread: I need help solving a trig proof

1. ## I need help solving a trig proof

I have been trying to solve this proof for hours and I just can't figure it out: (1+sinx+cosx)/(1+sinx-cosX)=(1+cosx)/sinx

2. Hi,

I'd do it this way :

\displaystyle \begin{aligned}\frac{1+\sin x+\cos x}{1+\sin x -\cos x}&= \frac{(1+\sin x+\cos x)(1+\sin x +\cos x)}{(1+\sin x -\cos x)(1+\sin x +\cos x)}\\&=\frac{(1+\sin x+\cos x)^2}{(1+\sin x -\cos x)(1+\sin x +\cos x)}\end{aligned}

Note that the denominator of the RHS looks like $\displaystyle (a-b)(a+b)=a^2-b^2$ with $\displaystyle a = 1+\sin x$ and $\displaystyle b=\cos x$ hence

\displaystyle \begin{aligned} \frac{1+\sin x+\cos x}{1+\sin x -\cos x}&=\frac{(1+\sin x)^2+2\cos x(1+\sin x)+\cos^2x}{(1+\sin x)^2-\cos^2x}\\ &=\frac{1+2\sin x+\sin^2x+2\cos x(1+\sin x)+\cos^2x}{1+2\sin x+\sin^2x-\cos^2x}\\ \end{aligned}

Using $\displaystyle 1-\cos^2x=\sin^2x$ to simplify the denominator and $\displaystyle \cos^2x=1-\sin^2x$ to simplify the numerator,

\displaystyle \begin{aligned} \frac{1+\sin x+\cos x}{1+\sin x -\cos x}&=\frac{1+2\sin x+\sin^2x+2\cos x(1+\sin x)+1-\sin^2x}{\sin^2x+2\sin x+\sin^2x}\\ &=\frac{2+2\sin x+2\cos x(1+\sin x)}{2\sin x(1+\sin x)}\\ &=\frac{2(1+\sin x)+2\cos x(1+\sin x)}{2\sin x(1+\sin x)}\\&=\frac{2(1+\sin x)(1+\cos x)}{2\sin x(1+\sin x)}\\ &=\frac{1+\cos x}{\sin x} \end{aligned}