Simplifying $\tan100^{\circ}+4\sin100^{\circ}$ The answer is $-\sqrt3$. 
I was wondering if this is just a coincidence?

Also, is there a relation between $$\tan(100^{\circ}+20^{\circ})=\frac{\tan100^{\circ}+\tan20^{\circ}}{1-\tan100^{\circ}.\tan20^{\circ}}=-\sqrt3$$ and the given expression? Or is there a more elegant method of solving the question? 
 A: One has $$\tan 100^\circ + 4\sin 100^\circ = \frac{\sin 100^\circ + 2\sin 200^\circ}{\cos 100^\circ} = \frac{\sin 100^\circ - 2\sin 20^\circ}{\cos 100^\circ} = \frac{2\cos 60^\circ\sin 40^\circ - \sin 20^\circ}{\cos 100^\circ} = \frac{\sin 40^\circ - \sin 20^\circ}{\cos 100^\circ} = \frac{2\cos 30^\circ\sin 10^\circ }{\cos 100^\circ} = -\sqrt{3}.$$
A: For the first question, let $\alpha=\cos100^\circ$.  By standard trig formulae you get
$$8\alpha^3-6\alpha=2\cos300^\circ=1\ .$$
Now the square of your expression is
$$\eqalign{\sin^2100^\circ\frac{(1+4\cos100^\circ)^2}{\cos^2100^\circ}
  &=\frac{(1-\alpha^2)(1+4\alpha)^2}{\alpha^2}\cr
  &=\frac{1+8\alpha+15\alpha^2-8\alpha^3-16\alpha^4}{\alpha^2}\cr
  &=\frac{8\alpha^3-6\alpha+8\alpha+15\alpha^2-8\alpha^3-16\alpha^4}{\alpha^2}\cr
  &=\frac{2+15\alpha-16\alpha^3}{\alpha}\cr
  &=\frac{16\alpha^3-12\alpha+15\alpha-16\alpha^3}{\alpha}\cr
  &=3\ .\cr}$$
It's not hard to see that your expression is negative, and hence it equals $-\sqrt3$.
A: Notice, $$\tan 100^\circ+4\sin 100^\circ$$
$$=\frac{\sin 100^\circ}{\cos 100^\circ}+4\sin 100^\circ$$
$$=\frac{\sin 100^\circ+4\sin 100^\circ\cos 100^\circ}{\cos 100^\circ}$$
$$=\frac{\sin 100^\circ+2\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{(\sin 200^\circ+\sin100^\circ)+\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{2\sin \left(\frac{200^\circ+100^\circ}{2}\right)\cos \left(\frac{200^\circ-100^\circ}{2}\right)+\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{2\sin 150^\circ \cos 50^\circ +\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{2\frac{1}{2} \cos 50^\circ +\sin (270^\circ-70^\circ)}{\cos (90^\circ+10^\circ)}$$
$$=\frac{\cos 50^\circ -\cos 70^\circ}{-\sin 10^\circ}=\frac{\cos 70^\circ -\cos 50^\circ}{\sin 10^\circ}$$
$$=\frac{2\sin \left(\frac{70^\circ+50^\circ}{2}\right)\sin \left(\frac{50^\circ-70^\circ}{2}\right)}{\sin 10^\circ}$$
$$=\frac{2\sin 60^\circ(-\sin 10^\circ)}{\sin 10^\circ}=-2\sin 60^\circ=-2\times \frac{\sqrt 3}{2}=\color{red}{-\sqrt 3}$$
A: If $$2\sec A\sin x+\tan x=-\tan A$$
Multiplying by $\cos A\cos x,$  $$2\sin x\cos x+\sin x\cos A=-\sin A\cos x$$
$$\sin(x+A)=-\sin2x=\sin(-2x)$$
$\implies$ either $(i), x+A=360^\circ n+(-2x)\iff x=120^\circ n-\dfrac A3$ where $n$ is any integer
In this problem $A=60^\circ$ and $n=1$
or $(ii), x+A=360^\circ n+180^\circ-(-2x)\iff x=A-n 360^\circ$ 
Related :
Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $
Prove that $ \tan40° + \sqrt 3 =4 \sin40° $
Trigonometry Simplification
Distant cousins:
Expressing a number in $\sqrt a/b$ form
Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$
