Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \tan \left( {x - \pi } \right).\cos \left( {x - \frac{\pi }{2}} \right).\cos \left( {\pi + x} \right)\\
= \tan x.\sin \left( {\frac{\pi }{2} - \left( {x - \frac{\pi }{2}} \right)} \right).\left[ { - \cos \left( {\pi - \left( {\pi + x} \right)} \right)} \right]\\
= \frac{{\sin x}}{{\cos x}}.sin\left( {\pi - x} \right).\left( { - \cos \left( { - x} \right)} \right)\\
= \frac{{\sin x}}{{\cos x}}.\sin x. - \cos x\\
= - {\sin ^2}x\\
b,\\
\cot \left( {\frac{\pi }{2} - x} \right) = \frac{{\cos \left( {\frac{\pi }{2} - x} \right)}}{{\sin \left( {\frac{\pi }{2} - x} \right)}} = \frac{{\sin x}}{{\cos x}} = \tan x\\
A = \frac{{{{\cos }^2}\left( {\frac{\pi }{2} - x} \right).{{\cos }^2}x + \sin \left( {\frac{\pi }{2} - x} \right).\cos \left( {\pi - x} \right)}}{{{{\sin }^2}x - {{\cos }^2}x}}\\
= \frac{{{{\tan }^2}x.{{\cos }^2}x + \cos x.\left( { - \cos x} \right)}}{{{{\sin }^2}x - {{\cos }^2}x}}\\
= \frac{{\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}.{{\cos }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x - {{\cos }^2}x}}\\
= \frac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x - {{\cos }^2}x}} = 1
\end{array}\)
