cho alpha là góc nhọn. cmr 1-tan a/1+tan a=cos a-sin a/cos a+sin a 16/09/2021 Bởi Harper cho alpha là góc nhọn. cmr 1-tan a/1+tan a=cos a-sin a/cos a+sin a
$VT=\dfrac{1-\tan a}{1+\tan a}$ $=\dfrac{1-\frac{\sin a}{\cos a}}{1+\frac{\sin a}{\cos a}}$ $=\dfrac{\cos a – \sin a}{\cos a} : \dfrac{\cos a + \sin a}{\cos a}$ $=\dfrac{\cos a -\sin a}{\cos a +\sin a}$ $=VP$ Bình luận
\[\frac{{1 – \tan \alpha }}{{1 + \tan \alpha }} = \frac{{1 – \frac{{\sin \alpha }}{{\cos \alpha }}}}{{1 + \frac{{\sin \alpha }}{{\cos \alpha }}}} = \frac{{\frac{{\cos \alpha – \sin \alpha }}{{\cos \alpha }}}}{{\frac{{\cos \alpha + \sin \alpha }}{{\cos \alpha }}}} = \frac{{\cos \alpha – \sin \alpha }}{{\cos \alpha + \sin \alpha }}\] Bình luận
$VT=\dfrac{1-\tan a}{1+\tan a}$
$=\dfrac{1-\frac{\sin a}{\cos a}}{1+\frac{\sin a}{\cos a}}$
$=\dfrac{\cos a – \sin a}{\cos a} : \dfrac{\cos a + \sin a}{\cos a}$
$=\dfrac{\cos a -\sin a}{\cos a +\sin a}$
$=VP$
\[\frac{{1 – \tan \alpha }}{{1 + \tan \alpha }} = \frac{{1 – \frac{{\sin \alpha }}{{\cos \alpha }}}}{{1 + \frac{{\sin \alpha }}{{\cos \alpha }}}} = \frac{{\frac{{\cos \alpha – \sin \alpha }}{{\cos \alpha }}}}{{\frac{{\cos \alpha + \sin \alpha }}{{\cos \alpha }}}} = \frac{{\cos \alpha – \sin \alpha }}{{\cos \alpha + \sin \alpha }}\]