Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{1 + 2\sin x.\cos x}} = \dfrac{{\left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right)}}{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 2\sin x.\cos x}}\\
= \dfrac{{\left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right)}}{{{{\left( {\sin x + \cos x} \right)}^2}}} = \dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}\\
= \dfrac{{\dfrac{{\sin x - \cos x}}{{\cos x}}}}{{\dfrac{{\sin x + \cos x}}{{\cos x}}}} = \dfrac{{\dfrac{{\sin x}}{{\cos x}} - 1}}{{\dfrac{{\sin x}}{{\cos x}} + 1}} = \dfrac{{\tan x - 1}}{{\tan x + 1}}\\
*)\\
3 - 4{\sin ^2}x = 3 - 4.\left( {1 - {{\cos }^2}x} \right) = 4{\cos ^2}x - 1\\
*)\\
\dfrac{{{{\sin }^2}a + 2{{\cos }^2}a - 1}}{{{{\cos }^2}a}} = \dfrac{{\left( {{{\sin }^2}a + {{\cos }^2}a} \right) - \left( {1 - {{\cos }^2}a} \right)}}{{{{\cos }^2}a}}\\
= \dfrac{{1 - {{\sin }^2}a}}{{{{\cos }^2}a}} = \dfrac{{{{\cos }^2}a}}{{{{\cos }^2}a}} = 1
\end{array}\)
