Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
P = \left( {{{\sin }^2}x - 1} \right).{\tan ^2}x + \left( {{{\cos }^2}x - 1} \right).{\cot ^2}x\\
= \left( { - {{\cos }^2}x} \right).\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \left( { - {{\sin }^2}x} \right).\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}\\
= - {\sin ^2}x - {\cos ^2}x\\
= - \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
= - 1\\
b,\\
P = \sin \left( {\dfrac{\pi }{2} + x} \right) + \cos \left( {x + \dfrac{{2\pi }}{3}} \right) + \cos \left( {x - \dfrac{{2\pi }}{3}} \right)\\
= \cos \left[ {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{2} + x} \right)} \right] + 2.\cos \dfrac{{x + \dfrac{{2\pi }}{3} + x - \dfrac{{2\pi }}{3}}}{2}.\cos \dfrac{{x + \dfrac{{2\pi }}{3} - x + \dfrac{{2\pi }}{3}}}{2}\\
= \cos \left( { - x} \right) + 2.\cos x.\cos \dfrac{{2\pi }}{3}\\
= \cos x + 2.\cos x.\left( { - \dfrac{1}{2}} \right)\\
= 0
\end{array}\)
