How to prove that a derivative of a formula equals to another formula. If $u= \ln(\tan x+\tan y+\tan z)$ prove $$\sin 2x \dfrac{du}{dx} + \sin 2y  \dfrac{du}{dy} + \sin 2z   \dfrac{du}{dz}=2 $$
My answwer was like this:
$$u' =\dfrac{ 1}{\tan x+\tan y+\tan z} \cdot( \sec^2 x +\sec^2 y +\sec^2 z )$$
$$   =\dfrac{ 1}{\dfrac{\sin x}{\cos x} + \dfrac{\sin y}{\cos y} + \dfrac{\sin z}{\cos z}}\cdot(\sec^2x + \sec^2y + \sec^2z)$$
$$  =\dfrac{ \cos x+\cos y+\cos z}{\sin x+\sin y+\sin z}\cdot(\sec^2x + \sec^2y + \sec^2z)$$
$$   = \sin 2x (\dfrac{\cos x}{\sin x})+ \sin 2y (\dfrac{\cos y}{\sin y}) + \sin 2z (\dfrac{\cos z}{\sin z})\cdot( \sec^2 x +\sec^2 y +\sec^2 z )$$
$$   =\dfrac{ 2\sin x\cos x}{\sin x} +\dfrac{ 2\sin y \cos y}{\sin x} +\dfrac{ 2\sin z\cos z}{\sin z}\cdot(\sec^2x + \sec^2y + \sec^2z)$$
$$=\dfrac{ \sin 2x}{\sin x} + \dfrac{ \sin 2y}{\sin x} + \dfrac{ \sin 2z}{\sin z}\cdot(\sec^2x + \sec^2y + \sec^2z)$$
I solved until here and I got stuck. Is my answer right until now? How could I finish it to prove that it equals to 2? Sorry for the wrong codes.

It seems like you should just compute the relevant partials and then verify that LHS $=$ RHS.
Something like:
$$\frac{\partial u}{\partial x} = \frac{1}{\tan x + \tan y + \tan z} \cdot \sec^2x \\
\frac{\partial u}{\partial y} = \frac{1}{\tan x + \tan y + \tan z} \cdot \sec^2y \\
\frac{\partial u}{\partial z} = \frac{1}{\tan x + \tan y + \tan z} \cdot \sec^2z \\$$ 
The LHS then becomes 
$$\frac{\sin(2x)\sec^2x + \sin(2y)\sec^2y+ \sin(2z)\sec^2z}{\tan x + \tan y + \tan z} = 2\left(\frac{ \tan x + \tan y + \tan z}{\tan x + \tan y + \tan z}\right) = 2$$
Which clearly equals the RHS.
