Show that Sin2A - tanA = (tanA).(cos2A) ??
I have so far worked it to :
Sin2A - tanA = - sinA/cosA + (2SinA).(cosA)
Really stuck now any help would be appreciated guys.
Show that Sin2A - tanA = (tanA).(cos2A) ??
I have so far worked it to :
Sin2A - tanA = - sinA/cosA + (2SinA).(cosA)
Really stuck now any help would be appreciated guys.
$\displaystyle \sin{2A} - \tan{A} = 2\sin{A}\cos{A} - \frac{\sin{A}}{\cos{A}}$
$\displaystyle = \frac{2\sin{A}\cos^2{A}}{\cos{A}} - \frac{\sin{A}}{\cos{A}}$
$\displaystyle = \frac{2\sin{A}\cos^2{A} - \sin{A}}{\cos{A}}$
$\displaystyle = \frac{\sin{A}(2\cos^2{A} - 1)}{\cos{A}}$
$\displaystyle = \left(\frac{\sin{A}}{\cos{A}}\right)(2\cos^2{A} - 1)$
$\displaystyle = \tan{A}\cos{2A}$.