CMR:(cosa-sina)/(cosa+sina)=(1/cos2)-tan2a 09/08/2021 Bởi Jasmine CMR:(cosa-sina)/(cosa+sina)=(1/cos2)-tan2a
$VT=\dfrac{\cos a-\sin a}{\cos a+\sin a}$ Nhân tử, mẫu với $\cos a-\sin a$: $=\dfrac{\cos^2a+\sin^2a-2\sin a\cos a}{(\cos a+\sin a)(\cos a-\sin a)}$ $=\dfrac{1-\sin 2a}{\cos 2a}$ $=\dfrac{1}{\cos 2a}-\tan 2a$ $=VP$ Bình luận
Đáp án: $\begin{array}{l}\dfrac{1}{{\cos 2a}} – \tan 2a\\ = \dfrac{1}{{\cos 2a}} – \dfrac{{\sin 2a}}{{\cos 2a}}\\ = \dfrac{{1 – \sin 2a}}{{\cos 2a}}\\ = \dfrac{{{{\cos }^2}a – 2.{\mathop{\rm sina}\nolimits} .cosa + {{\sin }^2}a}}{{{{\cos }^2}a – {{\sin }^2}a}}\\ = \dfrac{{{{\left( {\cos a – \sin a} \right)}^2}}}{{\left( {\cos a – \sin a} \right)\left( {\cos a + \sin a} \right)}}\\ = \dfrac{{\cos a – \sin a}}{{\cos a + \sin a}}\end{array}$ Bình luận
$VT=\dfrac{\cos a-\sin a}{\cos a+\sin a}$
Nhân tử, mẫu với $\cos a-\sin a$:
$=\dfrac{\cos^2a+\sin^2a-2\sin a\cos a}{(\cos a+\sin a)(\cos a-\sin a)}$
$=\dfrac{1-\sin 2a}{\cos 2a}$
$=\dfrac{1}{\cos 2a}-\tan 2a$
$=VP$
Đáp án:
$\begin{array}{l}
\dfrac{1}{{\cos 2a}} – \tan 2a\\
= \dfrac{1}{{\cos 2a}} – \dfrac{{\sin 2a}}{{\cos 2a}}\\
= \dfrac{{1 – \sin 2a}}{{\cos 2a}}\\
= \dfrac{{{{\cos }^2}a – 2.{\mathop{\rm sina}\nolimits} .cosa + {{\sin }^2}a}}{{{{\cos }^2}a – {{\sin }^2}a}}\\
= \dfrac{{{{\left( {\cos a – \sin a} \right)}^2}}}{{\left( {\cos a – \sin a} \right)\left( {\cos a + \sin a} \right)}}\\
= \dfrac{{\cos a – \sin a}}{{\cos a + \sin a}}
\end{array}$