CMR a) cot ² a – cos ² a = cot ² a . cos ² a b) $\frac{1 + cos a }{sin a}$ = $\frac{sin a}{1 – cos a}$ 09/07/2021 Bởi Remi CMR a) cot ² a – cos ² a = cot ² a . cos ² a b) $\frac{1 + cos a }{sin a}$ = $\frac{sin a}{1 – cos a}$
a) $\cot^2a – \cos^2a$ $= \dfrac{\cos^2a}{\sin^2a} – \cos^2a$ $= \dfrac{\cos^2a – \sin^2a.\cos^2a}{\sin^2a}$ $= \dfrac{\cos^2a(1 – \sin^2a)}{\sin^2a}$ $= \dfrac{\cos^2a.\cos^2a}{\sin^2a}$ $= \cot^2a.\cos^2a$ b) $\dfrac{1 + \cos a}{\sin a} = \dfrac{\sin a}{1 – \cos a}$ $\to (1 + \cos a)(1 – \cos a) = \sin^2a$ $\to 1 – \cos^2a = \sin^2a$ $\to \sin^2a = \sin^2a$ (hiển nhiên) Vậy $\dfrac{1 + \cos a}{\sin a} = \dfrac{\sin a}{1 – \cos a}$ Bình luận
a, $VT=\cot^2a-\cos^2a$ $=\dfrac{\cos^2a}{\sin^2a}-\cos^2a$ $=\dfrac{\cos^2a-\sin^2a\cos^2a}{\sin^2a}$ $=\dfrac{\cos^2a(1-\sin^2a)}{\sin^2a}$ $=\dfrac{\cos^2a.\cos^2a}{\sin^2a}$ $=\cot^2a.\cos^2a$ $=VP$ b, $VT=\dfrac{1+\cos a}{\sin a}$ $=\dfrac{(1+\cos a)(1-\cos a)}{\sin a(1-\cos a)}$ $=\dfrac{1-\cos^2a}{\sin a(1-\cos a)}$ $=\dfrac{\sin^2a}{\sin a(1-\cos a)}$ $=\dfrac{\sin a}{1-\cos a}$ $=VP$ Bình luận
a) $\cot^2a – \cos^2a$
$= \dfrac{\cos^2a}{\sin^2a} – \cos^2a$
$= \dfrac{\cos^2a – \sin^2a.\cos^2a}{\sin^2a}$
$= \dfrac{\cos^2a(1 – \sin^2a)}{\sin^2a}$
$= \dfrac{\cos^2a.\cos^2a}{\sin^2a}$
$= \cot^2a.\cos^2a$
b) $\dfrac{1 + \cos a}{\sin a} = \dfrac{\sin a}{1 – \cos a}$
$\to (1 + \cos a)(1 – \cos a) = \sin^2a$
$\to 1 – \cos^2a = \sin^2a$
$\to \sin^2a = \sin^2a$ (hiển nhiên)
Vậy $\dfrac{1 + \cos a}{\sin a} = \dfrac{\sin a}{1 – \cos a}$
a,
$VT=\cot^2a-\cos^2a$
$=\dfrac{\cos^2a}{\sin^2a}-\cos^2a$
$=\dfrac{\cos^2a-\sin^2a\cos^2a}{\sin^2a}$
$=\dfrac{\cos^2a(1-\sin^2a)}{\sin^2a}$
$=\dfrac{\cos^2a.\cos^2a}{\sin^2a}$
$=\cot^2a.\cos^2a$
$=VP$
b,
$VT=\dfrac{1+\cos a}{\sin a}$
$=\dfrac{(1+\cos a)(1-\cos a)}{\sin a(1-\cos a)}$
$=\dfrac{1-\cos^2a}{\sin a(1-\cos a)}$
$=\dfrac{\sin^2a}{\sin a(1-\cos a)}$
$=\dfrac{\sin a}{1-\cos a}$
$=VP$