$\begin{array}{l} a)\\ A = \dfrac{{\cos \left( {a + b} \right) + \sin a\sin b}}{{\cos \left( {a - b} \right) - \sin a\sin b}} = \dfrac{{\cos a\cos b - \sin a\sin b + \sin a\sin b}}{{\cos a\cos b + \sin a\sin b - \sin a\sin b}}\\ A = 1 \end{array}$
$\begin{array}{l} b)\\ B = \dfrac{{\sin \left( {a + b} \right) + \sin \left( {a - b} \right)}}{{\sin \left( {a + b} \right) - \sin \left( {a - b} \right)}} = \dfrac{{2\sin a\cos b}}{{2\cos a\sin b}} = \tan a.\cot b \end{array}$
$\begin{array}{l} c)\\ C = \dfrac{{2\cos \left( {\dfrac{\pi }{2} - x} \right)\sin \left( {\dfrac{\pi }{2} + x} \right)\tan \left( {\pi - x} \right)}}{{\cot \left( {\dfrac{\pi }{2} + x} \right)\sin \left( {\pi - x} \right)}} - 2\cos x\\ C = \dfrac{{2\sin x\cos \left( { - x} \right).\left( { - \tan x} \right)}}{{\tan \left( { - x} \right)\sin x}} - 2\cos x\\ C = \dfrac{{2\sin x\cos x\left( { - \tan x} \right)}}{{ - \tan x.\sin x}} - 2\cos x = 2\cos x - 2\cos x = 0 \end{array}$
