Chứng minh: $1 + \dfrac{1}{cos2\alpha} = \dfrac{tan2\alpha}{tan\alpha}$ 07/12/2021 Bởi Adeline Chứng minh: $1 + \dfrac{1}{cos2\alpha} = \dfrac{tan2\alpha}{tan\alpha}$
$VT=\dfrac{\cos^2a -\sin^2a+\sin^2a+\cos^2a}{\cos 2a}$ $=\dfrac{2\cos^2a}{\cos 2a}$ $VP=\dfrac{\sin 2a}{\cos 2a}.\dfrac{\cos a}{\sin a}$ $=\dfrac{2\sin a.\cos a.\cos a}{\cos 2a.\sin a}$ $=\dfrac{2\cos^2a}{\cos 2a}$ $= VP$ Bình luận
$\frac{tan2∝}{tan∝}$ $=\frac{sin2∝}{cos2∝}.\frac{cos∝}{sin∝}$ $=\frac{2cos^2∝}{cos2∝}$ $=\frac{2cos^2∝-1+1}{2cos^2∝-1}$ $=1+\frac{1}{2cos^2∝-1}$ $=1+\frac{1}{cos2∝}$ Bình luận
$VT=\dfrac{\cos^2a -\sin^2a+\sin^2a+\cos^2a}{\cos 2a}$
$=\dfrac{2\cos^2a}{\cos 2a}$
$VP=\dfrac{\sin 2a}{\cos 2a}.\dfrac{\cos a}{\sin a}$
$=\dfrac{2\sin a.\cos a.\cos a}{\cos 2a.\sin a}$
$=\dfrac{2\cos^2a}{\cos 2a}$
$= VP$
$\frac{tan2∝}{tan∝}$
$=\frac{sin2∝}{cos2∝}.\frac{cos∝}{sin∝}$
$=\frac{2cos^2∝}{cos2∝}$
$=\frac{2cos^2∝-1+1}{2cos^2∝-1}$
$=1+\frac{1}{2cos^2∝-1}$
$=1+\frac{1}{cos2∝}$