Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\cot a - \tan a = \dfrac{{\cos a}}{{\sin a}} - \dfrac{{\sin a}}{{\cos a}} \\= \dfrac{{{{\cos }^2}a - {{\sin }^2}a}}{{\sin a.\cos a}} = \dfrac{{\cos 2a}}{{\dfrac{1}{2}\sin 2a}} \\= 2\dfrac{{\cos 2a}}{{\sin 2a}} = 2\cot 2a\\
b,\\
\sin 2a.\left( {\tan a + \cot a} \right) = 2\sin a.\cos a.\left( {\dfrac{{\sin a}}{{\cos a}} + \dfrac{{\cos a}}{{\sin a}}} \right) \\= 2{\sin ^2}a + 2{\cos ^2}a\\ = 2\left( {{{\sin }^2}a + {{\cos }^2}a} \right) \\= 2.1 = 2\\
4,\\
\dfrac{{\tan x}}{{\sin x}} - \dfrac{{\sin x}}{{\cot x}} = \dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\sin x}} - \dfrac{{\sin x}}{{\dfrac{{\cos x}}{{\sin x}}}} \\= \dfrac{1}{{\cos x}} - \dfrac{{{{\sin }^2}x}}{{\cos x}} = \dfrac{{1 - {{\sin }^2}x}}{{\cos x}} = \dfrac{{{{\cos }^2}x}}{{\cos x}} = \cos x\\
5.\\
\dfrac{{1 + {{\sin }^2}x}}{{1 - {{\sin }^2}x}} = \dfrac{{\left( {1 - {{\sin }^2}x} \right) + 2{{\sin }^2}x}}{{1 - {{\sin }^2}x}} \\= 1 + \dfrac{{2{{\sin }^2}x}}{{1 - {{\sin }^2}x}} = 1 + \dfrac{{2{{\sin }^2}x}}{{{{\cos }^2}x}} \\= 1 + 2{\tan ^2}x\\
6,\\
\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cot }^2}x - {{\tan }^2}x}} = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}\\ = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\dfrac{{{{\cos }^4}x - {{\sin }^4}x}}{{{{\sin }^2}x.{{\cos }^2}x}}}} \\ ={\sin ^2}x.{\cos ^2}x.\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^4}x - {{\sin }^4}x}} \\= {\sin ^2}x.{\cos ^2}x.\dfrac{1}{{{{\sin }^2}x + {{\cos }^2}x}} \\= {\sin ^2}x.{\cos ^2}x.\dfrac{1}{1}\\ = {\sin ^2}x.{\cos ^2}x
\end{array}\)
