CHÚC BẠN HỌC TỐT!!!
Trả lời:
$a,(\sin x+\cos x)^2-(\sin x-\cos x)^2$
$=\sin^2x+2\sin x\cos x+\cos^2x-\sin^2x+2\sin x\cos x-\cos^2x$
$=1+2.2\sin x\cos x-1$
$=2\sin2x$
$b,VT=\dfrac{\sin x+\sin3x+\sin5x}{\cos x+\cos3x+\cos5x}$
$=\dfrac{(\sin x+\sin5x)+\sin3x}{(\cos x+\cos5x)+\cos3x}$
$=\dfrac{2\sin3x.\cos2x+\sin3x}{2\cos3x.\cos2x+\cos3x}$
$=\dfrac{\sin3x.(2\cos2x+1)}{\cos3x.(2\cos2x+1)}$
$=\dfrac{\sin3x}{\cos3x}$
$=\tan3x=VP$ (Đpcm).
