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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?

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One has $$\tan 100^\circ + 4\sin 100^\circ = \frac{\sin 100^\circ + 2\sin 200^\circ}{\cos 100^\circ} = \frac{\sin 100^\circ - 2\sin 20^\circ}{\cos 100^\circ} = \frac{2\cos 60^\circ\sin 40^\circ - \sin 20^\circ}{\cos 100^\circ} = \frac{\sin 40^\circ - \sin 20^\circ}{\cos 100^\circ} = \frac{2\cos 30^\circ\sin 10^\circ }{\cos 100^\circ} = -\sqrt{3}.$$

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    $\begingroup$ Now this is what I call the beauty of math! $100=90+10=60+40=120-20$! $\endgroup$ – Aditya Agarwal Sep 14 '15 at 5:16
  • $\begingroup$ Absolutely +1 for the sleek solution $\endgroup$ – Shailesh Sep 14 '15 at 13:59
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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 $-\sqrt3$.

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  • $\begingroup$ Nice way to do it. +1 $\endgroup$ – Shailesh Sep 14 '15 at 13:59
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Notice, $$\tan 100^\circ+4\sin 100^\circ$$ $$=\frac{\sin 100^\circ}{\cos 100^\circ}+4\sin 100^\circ$$ $$=\frac{\sin 100^\circ+4\sin 100^\circ\cos 100^\circ}{\cos 100^\circ}$$ $$=\frac{\sin 100^\circ+2\sin 200^\circ}{\cos 100^\circ}$$ $$=\frac{(\sin 200^\circ+\sin100^\circ)+\sin 200^\circ}{\cos 100^\circ}$$ $$=\frac{2\sin \left(\frac{200^\circ+100^\circ}{2}\right)\cos \left(\frac{200^\circ-100^\circ}{2}\right)+\sin 200^\circ}{\cos 100^\circ}$$ $$=\frac{2\sin 150^\circ \cos 50^\circ +\sin 200^\circ}{\cos 100^\circ}$$ $$=\frac{2\frac{1}{2} \cos 50^\circ +\sin (270^\circ-70^\circ)}{\cos (90^\circ+10^\circ)}$$ $$=\frac{\cos 50^\circ -\cos 70^\circ}{-\sin 10^\circ}=\frac{\cos 70^\circ -\cos 50^\circ}{\sin 10^\circ}$$ $$=\frac{2\sin \left(\frac{70^\circ+50^\circ}{2}\right)\sin \left(\frac{50^\circ-70^\circ}{2}\right)}{\sin 10^\circ}$$ $$=\frac{2\sin 60^\circ(-\sin 10^\circ)}{\sin 10^\circ}=-2\sin 60^\circ=-2\times \frac{\sqrt 3}{2}=\color{red}{-\sqrt 3}$$

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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 $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$

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