Chứng minh các đẳng thức lượng giác

$\begin{array}{l}
b)\tan {20^o} - \tan {40^o} + \tan {80^o}\\
 = \dfrac{{\sin {{20}^o}}}{{\cos {{20}^o}}} + \dfrac{{\sin {{80}^o}}}{{\cos {{80}^o}}} - \dfrac{{\sin {{40}^o}}}{{\cos {{40}^o}}}\\
 = \dfrac{{\sin {{20}^o}}}{{\cos {{20}^o}}} + \dfrac{{\sin {{80}^o}.\cos {{40}^o} - \sin {{40}^o}\cos {{80}^o}}}{{\cos {{80}^o}.\cos {{40}^o}}}\\
 = \dfrac{{\sin {{20}^o}}}{{\cos {{20}^o}}} + \dfrac{{\sin \left( {{{80}^o} - {{40}^o}} \right)}}{{\dfrac{1}{2}\left[ {\cos {{120}^o} + \cos {{40}^o}} \right]}}\\
 = \dfrac{{\sin {{20}^o}}}{{\cos {{20}^o}}} + \dfrac{{2\sin {{40}^o}}}{{\cos {{40}^o} - \dfrac{1}{2}}}\\
 = \dfrac{{\sin {{20}^o}}}{{\cos {{20}^o}}} + \dfrac{{4\sin {{40}^o}}}{{2\cos {{40}^o} - 1}}\\
 = \dfrac{{\sin {{20}^o}}}{{\cos {{20}^o}}} + \dfrac{{4\sin {{40}^o}}}{{2\left( {2{{\cos }^2}{{20}^o} - 1} \right) - 1}}\\
 = \dfrac{{\sin {{20}^o}}}{{\cos {{20}^o}}} + \dfrac{{4\sin {{40}^o}}}{{4{{\cos }^2}{{20}^o} - 3}}\\
 = \dfrac{{\sin {{20}^o}\left[ {4{{\cos }^2}{{20}^o} - 3} \right] + 4\sin {{40}^o}.\cos {{20}^o}}}{{4{{\cos }^3}{{20}^o} - 3\cos {{20}^o}}}\\
 = \dfrac{{4{{\cos }^2}{{20}^o}.\sin {{20}^o} + 4\sin {{40}^o}.\cos {{20}^o} - 3\sin {{20}^o}}}{{\cos \left( {{{3.20}^o}} \right)}}\\
 = \dfrac{{2\sin {{40}^o}.\cos {{20}^o} + 4\sin {{40}^o}.\cos {{20}^o} - 3\sin {{20}^o}}}{{\cos {{60}^o}}}\\
 = \dfrac{{6\sin {{40}^o}.\cos {{20}^o} - 3\sin {{20}^o}}}{{\cos {{60}^o}}}\\
 = \dfrac{{3\sin {{60}^o} + 3\sin {{20}^o} - 3\sin {{20}^o}}}{{\cos {{60}^o}}}\\
 = \dfrac{{3\sin {{60}^o}}}{{\cos {{60}^o}}} = 3\tan {60^o} = 3\sqrt 3 \\
c)\tan {10^o} - \tan {50^o} + \tan {70^o} + \tan {60^o}\\
 = \dfrac{{\sin {{10}^o}}}{{\cos {{10}^o}}} + \dfrac{{\sin {{70}^o}}}{{\cos {{70}^o}}} - \dfrac{{\sin {{50}^o}}}{{\cos {{50}^o}}} + \tan {60^o}\\
 = \dfrac{{\sin {{10}^o}}}{{\cos {{10}^o}}} + \dfrac{{\sin {{70}^o}\cos {{50}^o} - \sin {{50}^o}\cos {{70}^o}}}{{\cos {{70}^o}.\cos {{50}^o}}} + \tan {60^o}\\
 = \dfrac{{\sin {{10}^o}}}{{\cos {{10}^o}}} + \dfrac{{\sin \left( {{{70}^o} - {{50}^o}} \right)}}{{\dfrac{1}{2}\left( {\cos {{120}^o} + \cos {{20}^o}} \right)}} + \tan {60^o}\\
 = \dfrac{{\sin {{10}^o}}}{{\cos {{10}^o}}} + \dfrac{{2\sin {{20}^o}}}{{\cos {{20}^o} - \dfrac{1}{2}}} + \tan {60^o}\\
 = \dfrac{{\sin {{10}^o}}}{{\cos {{10}^o}}} + \dfrac{{4\sin {{20}^o}}}{{2\cos {{20}^o} - 1}} + \tan {60^o}\\
 = \dfrac{{\sin {{10}^o}}}{{\cos {{10}^o}}} + \dfrac{{4\sin {{20}^o}}}{{2\left( {2{{\cos }^2}{{10}^o} - 1} \right) - 1}} + \tan {60^o}\\
 = \dfrac{{\sin {{10}^o}}}{{\cos {{10}^o}}} + \dfrac{{4\sin {{20}^o}}}{{4{{\cos }^2}{{10}^o} - 3}} + \tan {60^o}\\
 = \dfrac{{4{{\cos }^2}{{10}^o}.\sin {{10}^o} - 3\sin {{10}^o} + 4\sin {{20}^o}\cos {{10}^o}}}{{4{{\cos }^3}{{10}^o} - 3\cos {{10}^o}}} + \tan {60^o}\\
 = \dfrac{{2\sin {{20}^o}.\cos {{10}^o} + 4\sin {{20}^o}.\cos {{10}^o} - 3\sin {{10}^o}}}{{\cos \left( {{{30}^o}} \right)}} + \tan {60^o}\\
 = \dfrac{{6\sin {{20}^o}\cos {{10}^o} - 3\sin {{10}^o}}}{{\cos \left( {{{30}^o}} \right)}} + \tan {60^o}\\
 = \dfrac{{3\sin {{30}^o} + 3\sin {{10}^o} - 3\sin {{10}^o}}}{{\cos \left( {{{30}^o}} \right)}} + \tan {60^o}\\
 = \dfrac{{3\sin {{30}^o}}}{{\cos {{30}^o}}} + \tan {60^o}\\
 = 3\tan {30^o} + \tan {60^o} = 3.\dfrac{{\sqrt 3 }}{3} + \sqrt 3  = 2\sqrt 3 
\end{array}$