Đáp án:
điều phải chứng minh
Giải thích các bước giải:
\(\begin{array}{l}
b)VT = \dfrac{{1 + 2\sin a\cos a}}{{{{\sin }^2}a - {{\cos }^2}a}} = \dfrac{{{{\sin }^2}a + {{\cos }^2}a + 2\sin a\cos a}}{{{{\sin }^2}a - {{\cos }^2}a}}\\
= \dfrac{{{{\left( {\sin a + \cos a} \right)}^2}}}{{\left( {\sin a + \cos a} \right)\left( {\sin a - \cos a} \right)}}\\
= \dfrac{{\sin a + \cos a}}{{\sin a - \cos a}}\\
VP = \left( {\dfrac{{\sin a}}{{\cos a}} + 1} \right):\left( {\dfrac{{\sin a}}{{\cos a}} - 1} \right)\\
= \dfrac{{\sin a + \cos a}}{{\cos a}}:\dfrac{{\sin a - \cos a}}{{\cos a}}\\
= \dfrac{{\sin a + \cos a}}{{\sin a - \cos a}}\\
\to VT = VP\\
\to dpcm\\
c)VP = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{\sin x}}{{\cos x}}.\dfrac{{\cos x}}{{\sin x}}\\
= {\tan ^2}x + 1 = \dfrac{1}{{{{\cos }^2}x}} = VT\\
\to dpcm\\
f)VT = {\sin ^4}x + {\cos ^4}x + 2{\sin ^2}x{\cos ^2}x - 2{\sin ^2}x{\cos ^2}x\\
= {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\\
= 1 - 2{\sin ^2}x{\cos ^2}x = VP\\
\to dpcm
\end{array}\)
