a)$\begin{array}{l} \dfrac{1}{{\sin {{10}^o}}} - 4\sin {70^o} = \dfrac{1}{{\sin {{10}^o}}} - 4\sin \left( {{{60}^o} + {{10}^o}} \right)\\  = \dfrac{1}{{\sin {{10}^o}}} - 4\left( {\sin {{60}^o}\cos {{10}^o} + \sin {{10}^o}.\cos {{60}^o}} \right)\\  = \dfrac{1}{{\sin {{10}^o}}} - 4\left( {\dfrac{{\sqrt 3 }}{2}.\cos {{10}^o} + \sin {{10}^o}.\dfrac{1}{2}} \right)\\  = \dfrac{1}{{\sin {{10}^o}}} - 2\sqrt 3 .\cos {10^o} - 2\sin {10^o}\\  = \dfrac{{1 - 2{{\sin }^2}{{10}^o} - 2\sqrt 3 .\cos {{10}^o}.\sin {{10}^o}}}{{\sin {{10}^o}}}\\  = \dfrac{{\cos {{20}^o} - \sqrt 3 \sin {{20}^o}}}{{\sin {{10}^o}}} = \dfrac{{\dfrac{1}{2}.\cos {{20}^o} - \dfrac{{\sqrt 3 }}{2}.\sin {{20}^o}}}{{\dfrac{1}{2}.\sin {{10}^o}}}\\  = \dfrac{{\cos {{60}^o}.\cos {{20}^o} - \sin {{60}^o}.\sin {{20}^o}}}{{\dfrac{1}{2}.\sin {{10}^o}}} = \dfrac{{\cos \left( {{{60}^o} + {{20}^o}} \right)}}{{\dfrac{1}{2}.\sin {{10}^o}}}\\  = \dfrac{{\cos {{80}^o}}}{{\dfrac{1}{2}.\sin {{10}^o}}} = \dfrac{{\sin {{10}^o}}}{{\dfrac{1}{2}.\sin {{10}^o}}} = 2 \end{array}$

b)$\begin{array}{l} B = \cos {14^o} + \cos {134^o} + \cos {106^o}\\ B = \cos {14^o} + \cos \left( {{{120}^o} + {{14}^o}} \right) + \cos \left( {{{120}^o} - {{16}^o}} \right)\\ B = \cos {14^o} + \cos {120^o}.\cos {14^o} - \sin {120^o}.\sin {14^o} + \cos {120^o}.\cos {14^o} + \sin {120^o}.\sin {14^o}\\ B = \cos {14^o} + 2\cos {120^o}.\cos {14^o} = \cos {14^o} - 2.\dfrac{1}{2}.\cos {14^o} = 0 \end{array}$

c)$\begin{array}{l} C = {\sin ^4}\dfrac{\pi }{{16}} + {\sin ^4}\dfrac{{3\pi }}{{16}} + {\sin ^4}\dfrac{{5\pi }}{{16}} + {\sin ^4}\dfrac{{7\pi }}{{16}}\\ C = \left( {{{\sin }^4}\dfrac{\pi }{{16}} + {{\sin }^4}\dfrac{{7\pi }}{{16}}} \right) + \left( {{{\sin }^4}\dfrac{{3\pi }}{{16}} + {{\sin }^4}\dfrac{{5\pi }}{{16}}} \right)\\ C = \left( {{{\sin }^4}\dfrac{\pi }{{16}} + {{\left( {{{\cos }^2}\dfrac{\pi }{{16}}} \right)}^2}} \right) + \left( {{{\sin }^4}\dfrac{{3\pi }}{{16}} + {{\left( {{{\cos }^2}\dfrac{{3\pi }}{{16}}} \right)}^2}} \right)\\ C = {\sin ^4}\dfrac{\pi }{{16}} + {\cos ^4}\dfrac{\pi }{{16}} + {\sin ^4}\dfrac{{3\pi }}{{16}} + {\cos ^4}\dfrac{{3\pi }}{{16}}\\ C = {\left( {{{\sin }^2}\dfrac{\pi }{{16}} + {{\cos }^2}\dfrac{\pi }{{16}}} \right)^2} + {\left( {{{\sin }^2}\dfrac{{3\pi }}{{16}} + {{\cos }^2}\dfrac{{3\pi }}{{16}}} \right)^2} - 2{\sin ^2}\dfrac{\pi }{{16}}.{\cos ^2}\dfrac{\pi }{{16}} - 2{\sin ^2}\dfrac{{3\pi }}{{16}}.{\cos ^2}\dfrac{{3\pi }}{{16}}\\ C = 2 - \dfrac{1}{2}.{\sin ^2}\dfrac{\pi }{8} - \dfrac{1}{2}{\sin ^2}\dfrac{{3\pi }}{8}\\ C = 2 - \dfrac{1}{2}\left( {{{\sin }^2}\dfrac{\pi }{8} + {{\sin }^2}\dfrac{{3\pi }}{8}} \right)\\ C = 2 - \dfrac{1}{2}\left( {{{\sin }^2}\dfrac{\pi }{8} + {{\cos }^2}\dfrac{\pi }{8}} \right)\\ C = \dfrac{3}{2} \end{array}$

d)$\begin{array}{l} CMR:\cot x - \tan x = \dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}} = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x.\cos x}} = \dfrac{{\cos 2x}}{{\dfrac{1}{2}.\sin 2x}} = 2\cot 2x\\ D = \cot 7,{5^o} + \tan 67,{5^o} - \tan 7,{5^o} - \cot 67,{5^o}\\ D = \cot 7,{5^o} - \tan 7,{5^o} - \left( {\cot 67,{5^o} - \tan 67,{5^o}} \right)\\ D = 2\cot {15^o} - 2\cot {135^o} = 2\left( {2 + \sqrt 3 } \right) + 2 = 6 + 2\sqrt 3  \end{array}$