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