Find the general solution of equation \cot x+\tan x=2 My approach: \cot

Let P(x, y) be the terminal point on the unit circle determined by t. Then $\sin t=,\cos t=,\text{and}\tan t=$.

Is ${\displaystyle \arccos \left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)=\arctan \left(\frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{6}}\right)}$?
They are equal according to the calculator but how?
I made a triangle with base ${\displaystyle \sqrt{6}+1}$ and hypotenuse ${\displaystyle 2\sqrt{3}}$
Then the height comes out to be ${\displaystyle \sqrt{5-2\sqrt{6}}}$ using Pythagoras theorem.
But it should come out to be ${\displaystyle \sqrt{3}-\sqrt{2}}$

Prove the Identity - Double Angle
${\displaystyle \cot \left(A\right)=\frac{\sin \left(2A\right)}{1-\cos \left(2A\right)}}$

Solving
${\displaystyle E=\frac{1}{{\sin 10}^{\circ }}-\frac{\sqrt{3}}{{\cos 10}^{\circ }}}$

Evaluating ${\displaystyle \sin (\arccos \frac{1}{2}+\arccos \frac{7}{25})}$

A problem in trigonometry
If ${\displaystyle \tan (x+y)=a+b}$ and ${\displaystyle \tan (x-y)=a-b}$ Then prove that: ${\displaystyle a\tan \left(x\right)-b\tan \left(y\right)=a^2-b^2}$
I have used the formulae for ${\displaystyle \tan (x+y)}$ and ${\displaystyle \tan (x-y)}$ and then cross multiplied and then found a and b individually and then tried to put the values of a and b in "${\displaystyle a\tan \left(x\right)-b\tan \left(y\right)}$"(l.h.s) but it did not lead to the r.h.s.

I don't find the right identities for this
${\displaystyle \underset{x\to \frac{\pi }{4}}{lim}\left(\tan \left(2x\right)\tan (\frac{\pi }{4}-x)\right)}$