Prove the identity $\frac{1}{\tan (x)(1+\cos( 2x))} = \csc(2x)$ $$\frac{1}{\tan (x)(1+\cos(2x))} = \csc(2x)$$
I really don't know what to do with denominator. Sure, I can use the double angle formula for cosine, and get:
$$\frac{1}{\tan(x)(2 - 2\sin^2(x))} = \csc(2x)$$
But what's next?
 A: $$\frac{1}{\color{#0F0}{\tan x}(\color{#00F}{1}+\color{#F00}{\cos 2x)}}=\frac{\color{#0f0}{\cos x}}{\color{#0f0}{\sin x}(\color{#00F}{\sin^2 x+\cos^2 x}+\color{#F00}{\cos^2x-\sin^2 x})}$$
$$=\frac{1}{2\sin x\cos x}=\frac{1}{\sin 2x}=\csc 2x$$
A: There's a better form of the double-angle formula for this purpose: remember there are three different forms of the cosine double-angle formula,
$$ \cos{2x} = \cos^2{x}-\sin^2{x} = 2\cos^2{x}-1=1-2\sin^2{x}. $$
Hence
$$ 1+\cos{2x} = 2\cos^2{x}. $$
Now, you also know $\tan{x}=\frac{\sin{x}}{\cos{x}}$ (right?), so you have
$$ \frac{1}{\tan{x}(1+\cos{2x})} = \frac{\cos{x}}{\sin{x} \cdot 2\cos^2{x}} = \frac{1}{2\sin{x}\cos{x}} = \frac{1}{\sin{2x}} = \csc{2x}, $$
using the sine double-angle formula.
A: We have to prove $$\frac{1}{\tan(x)(1+\cos(2x))}=\csc(2x)=\frac{1}{\sin(2x)}$$ Multiply both sides by $\tan(x)$ and apply $\sin(2x)=2\sin(x)\cos(x)$ and you arrive at $$\frac{1}{2{\cos^2(x)}}=\frac{1}{1+\cos(2x)}$$ raise both sides to the power of -1, and divide both sides by two and arrive at $$\cos^2(x)=\frac{1}{2}+\frac{1}{2}\cos(2x)$$ which is a well-known result that can be easily derived using Euler's formula.
A: Here's one more -- this is being a bit "cute", but it is related to another trigonometric relation you should be meeting soon, if you haven't already.
$$ \frac{1}{\tan (x)(1+\cos(2x))}  \ = \  \csc(2x) \ \ \Rightarrow \ \ \frac{1}{\tan (x)(1+\cos(2x))}  \ = \  \frac{1}{\sin(2x)} $$
$$ \Rightarrow \ \ \tan (x) \ = \ \frac{\sin(2x)}{1+\cos(2x)}  \ \ . $$
If we now identify $ \ 2 \ x \ $ as the angle $ \ \theta \ $ , then $ \ x \ = \ \frac{\theta}{2} \ $ and we have
$$ \Rightarrow \ \ \tan \left(\frac{\theta}{2}\right) \ = \ \frac{\sin(\theta)}{1+\cos(\theta)}  \ \ , $$
which is one form of the "half-angle formula" for the tangent function.  To prove your original relation, you can start from this equation and work backwards.
