Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
0 < a < \frac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a > 0
\end{array} \right.\\
\sin a > 0 \Rightarrow \sin a = \sqrt {1 - {{\cos }^2}a} = \frac{4}{5}\\
\sin 2a = 2.\sin a.\cos a = 2.\frac{3}{5}.\frac{4}{5} = \frac{{24}}{{25}}\\
\cos 2a = 2{\cos ^2}a - 1 = \frac{{ - 7}}{{25}}\\
\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - 24}}{7}\\
b,\\
\frac{{{{\left( {\sin x + \cos x} \right)}^2} - 1}}{{\cot x - \sin x.\cos x}} = \frac{{{{\sin }^2}x + 2\sin x.\cos x + {{\cos }^2}x - 1}}{{\frac{{\cos x}}{{\sin x}} - \sin x.\cos x}}\\
= \frac{{2\sin x.\cos x}}{{\frac{{\cos x - {{\sin }^2}x.\cos x}}{{\sin x}}}} = \frac{{2{{\sin }^2}x.\cos x}}{{\cos x - {{\sin }^2}x.\cos x}} = \frac{{2{{\sin }^2}x}}{{1 - {{\sin }^2}x}} = \frac{{2{{\sin }^2}x}}{{{{\cos }^2}x}} = 2{\tan ^2}x\\
c,\\
\cot x = 1 \Leftrightarrow \frac{{\cos x}}{{\sin x}} = 1 \Leftrightarrow \sin x = \cos x\\
B = \frac{{7{{\sin }^2}x - 2\cos x.\sin x + 5{{\cos }^2}x}}{{3{{\cos }^2}x + 2\sin x.\cos x - 4{{\sin }^2}x}}\\
= \frac{{7{{\sin }^2}x - 2\sin x.\sin x + 5{{\sin }^2}x}}{{3{{\sin }^2}x + 2.\sin x.\sin x - 4{{\sin }^2}x}}\\
= \frac{{10{{\sin }^2}x}}{{{{\sin }^2}x}} = 10\\
d,\\
B = \frac{{1 + \sin x}}{{\cos x}}.\left( {1 - \frac{{{{\left( {1 - \sin x} \right)}^2}}}{{{{\cos }^2}x}}} \right)\\
= \frac{{1 + \sin x}}{{\cos x}}.\frac{{{{\cos }^2}x - {{\left( {1 - \sin x} \right)}^2}}}{{{{\cos }^2}x}}\\
= \frac{{1 + \sin x}}{{\cos x}}.\frac{{{{\cos }^2}x - 1 + 2\sin x - {{\sin }^2}x}}{{{{\cos }^2}x}}\\
= \frac{{1 + \sin x}}{{\cos x}}.\frac{{ - {{\sin }^2}x + 2\sin x - {{\sin }^2}x}}{{{{\cos }^2}x}}\\
= \frac{{1 + \sin x}}{{\cos x}}.\frac{{2\sin x\left( {1 - \sin x} \right)}}{{{{\cos }^2}x}}\\
= \frac{{2\sin x.\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}{{{{\cos }^3}x}}\\
= \frac{{2.\sin x.\left( {1 - {{\sin }^2}x} \right)}}{{{{\cos }^3}x}}\\
= \frac{{2\sin x.{{\cos }^2}x}}{{{{\cos }^3}x}} = \frac{{2\sin x}}{{\cos x}} = 2\tan x
\end{array}\)
