How to find $\theta$ ?? If $0\leq\theta\leq360$, and:
$$\tan \theta=2+\sqrt{3}$$
 Where $\theta$ is in degrees .
How to find $\theta$  (without using a calculator) ? Can anyone give me some hints?
 A: HINT:
$$2+\sqrt3=\tan(45^\circ+y)=\dfrac{1+\tan y}{1-\tan y}$$
$$\iff\tan y=\dfrac{2+\sqrt3-1}{2+\sqrt3+1}=\dfrac1{\sqrt3}=\tan ?^\circ$$
A: Double the angle "to see", and get
$$\tan(2\theta)=\frac{2(2+\sqrt3)}{1-(2+\sqrt3)^2}=-\frac1{\sqrt3}=-\tan(30°).$$
Then $\theta$ must be one of $75°,165°,255°,345°$, of which two angles must be rejected (they were introduced by the doubling). A quick look at the trigonometric circle shows that the solutions belong to the first or third quadrant,
$$75°,255°.$$

Actually, this is a general and systematic approach: it will work for any tangent of a rational number of half-turns, $$\theta=\frac pq180°.$$
You can try all integers $q$ in turn and compute $\tan(q\theta)$ by successive applications of the addition formula, until you get to $\tan(p180°)=0$ (but you can stop earlier if you get to a known value).
For example, from
$$\tan(\theta)=\sqrt{5-2\sqrt{5}}$$ you get
$$\begin{align}\tan(2\theta)&=(\sqrt5+2)\sqrt{5-2\sqrt{5}},\\
\tan(3\theta)&=-\frac14(\sqrt5+1)(\sqrt5+3)\sqrt{5-2\sqrt{5}},\\
\tan(4\theta)&=-\sqrt{5-2\sqrt{5}},\\
\tan(5\theta)&=0.\end{align}$$
A: The given tangent is not one of the well-known values, so you need to look for a combination of known values, probably by the angle addition/subtraction formula.
The term $\sqrt3$ hints you to look for the angle $60°$. So try
$$\tan(\theta-60°)=\frac{\tan(\theta)-\tan(60°)}{1+\tan(\theta)\tan(60°)}=\frac{2+\sqrt3-\sqrt3}{1+(2+\sqrt3)\sqrt3}=\frac1{2+\sqrt3}.$$
Bingo, you see that this is the inverse of the given $\tan(\theta)$, or $\cot(\theta)=\tan(90°-\theta)$. Then
$$\theta-60°=90°-\theta,$$ or
$$\theta=75°.$$
Note that
$$\theta=75°+180°=255°$$ is also a solution.
A: We have 
$$\tan(75°)=\tan(30°+45°)=\frac{\tan(30°)+\tan(45)°}{1-\tan(30°)\tan(45°)}=$$ $$\frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}}=\frac{\sqrt{3}+3}{3-\sqrt{3}}
=\frac{(3+\sqrt{3})(3+\sqrt{3})}{9-3}=\frac{12+6\sqrt{3}}{6}=2+\sqrt{3}$$
So, one solution is $75°$ . The other is $75°+180°=255°$
A: $$\sec2A+\tan2A=\dfrac{1+\sin2A}{\cos2A}=\tan\left(45^\circ+A\right)$$
Here $A=30^\circ$
