Bài 7:
$B = \cos^2\alpha + \cos^2\alpha\sin^2\alpha + \sin^4\alpha$
$= \cos^2\alpha + \sin^2\alpha(\cos^2\alpha + \sin^2\alpha)$
$= \cos^2\alpha + \sin^2\alpha.1$
$= 1$
$C = \dfrac{1}{1 + \sin\alpha} + \dfrac{1}{1 - \sin\alpha} - 2\tan^2\alpha$
$= \dfrac{1 - \sin\alpha + 1 + \sin\alpha}{(1 + \sin\alpha)(1 - \sin\alpha)} - 2\tan^2\alpha$
$=\dfrac{2}{1 - \sin^2\alpha} -2\tan^2\alpha$
$= \dfrac{2}{\cos^2\alpha} - 2\dfrac{\sin^2\alpha}{\cos^2\alpha}$
$= \dfrac{2 - 2\sin^2\alpha}{\cos^2\alpha}$
$= \dfrac{2(1 - \sin^2\alpha)}{\cos^2\alpha}$
$= 2\dfrac{\cos^2\alpha}{\cos^2\alpha}$
$=2$
Bài 8: Sửa đề: $\sin2\alpha = 2\sin\alpha.\cos\alpha$
Ta có: $AM$ là trung tuyến ứng với cạnh huyền $BC$
$\Rightarrow AM =MC = MB = \dfrac{BC}{2}$
$\Rightarrow ΔMAC$ cân tại $M$
$\Rightarrow \widehat{ACB} = \widehat{ACM} = \widehat{MAC} = \alpha$
$\Rightarrow \widehat{HMA} = 2\widehat{ACB} = 2\alpha$
Ta được:
$\sin2\alpha = \sin\widehat{HMA} = \dfrac{AH}{AM} = \dfrac{2AH}{BC} = \dfrac{2AH.BC}{BC.BC}$
$= \dfrac{2.AB.AC}{BC.BC} = 2.\dfrac{AB}{BC}\cdot \dfrac{AC}{BC} = 2.\sin\widehat{ACB}.\cos\widehat{ACB}$
$= 2\sin\alpha.\cos\alpha$
