Can you solve this without Calculator Attack?

$(\sin\quad 40)(\cot\quad 20+\tan\quad 20)-2\\ =(\sin\quad 40)(\cfrac { \cos\quad 20 }{ \sin\quad 20 } +\cfrac { \sin\quad 20 }{ \cos\quad 20 } )-2\\ =(\sin\quad 40)(\cfrac { { \cos }^{ 2 }\quad 20+\sin^{ 2 }\quad 20 }{ \sin\quad 20\quad \cos\quad 20 } )-2\\ =(\sin\quad 40)(\cfrac { 1 }{\sin\quad 20\quad \cos\quad 20 } )-2\\ =(\sin\quad 40)(\cfrac { 2 }{ 2\sin\quad 20\quad \cos\quad 20 } )-2\\ =(\sin\quad 40\times \cfrac { 2 }{ \sin\quad 40 } )-2\\ =2-2\\=0$

It is known that $\space \sin{\theta} = \dfrac {2\tan{\frac{\theta}{2}}}{1+\tan^2 {\frac{\theta}{2}}}$ . Therefore,

$\begin{aligned} \sin{40^\circ}(\cot{20^\circ}+\tan{20^\circ} )-2 & = \dfrac {2 \tan{20^\circ}}{1+\tan^2{20^\circ}} \left( \dfrac {1}{\tan{20^\circ}} + \tan{20^\circ} \right) - 2 \\ & = \dfrac {2 \tan{20^\circ}}{1+\tan^2{20^\circ}} \left( \dfrac {1+\tan^2{20^\circ}}{\tan{20^\circ}} \right) - 2 \\ & = 2 - 2 = \boxed{0} \end{aligned}$

correct Paul. did the same way..

$(\sin40) ( \cot 20 + \tan20) - 2$

$\Rightarrow (2\sin20\cos20) ( \cot 20 + \tan20) - 2$

$\Rightarrow 2\sin20\cos20 (\frac{cos 20}{sin20}) + 2\sin20\cos20 (\frac{sin20}{cos20}) - 2$

$\Rightarrow 2cos^220 + 2sin^220 -2$

$\Rightarrow 2(cos^220 + sin^220)-2 = 2(1)-2 = 0$