Chủ đề này có 1 trả lời

[sakura139]

Tính:

$tan \ 20^o.tan \ 80^o + tan \ 80^o.tan \ 140^o + tan \ 140^o.tan20^o$

[lovethislife1997]

Đặt $t=80^{\circ}$

Biểu thức ban đầu$=tan(x-60^{\circ}).tanx+tanx.tan(x+60^{\circ})+tan(x-60^{\circ}).tan(x+60^{\circ})$

$=tanx[tan(x-60^{\circ})+tan(x+60^{\circ})]+tan(x-60^{\circ}).tan(x+60^{\circ})$

$=\frac{sinx}{cosx}.\frac{sin(x+60^{\circ}+x-60^{\circ})}{cos(x+60^{\circ}).cos(x-60^{\circ})}+\frac{sin(x-60^{\circ}).sin(x+60^{\circ})}{cos(x-60^{\circ}).cos(x+60^{\circ})}$

$=\frac{sinx}{cosx}.\frac{sin2x}{cos(x+60^{\circ}).cos(x-60^{\circ})}+\frac{sin^2x.cos^260-sin^260.cos^2x}{cos(x+60^{\circ}).cos(x-60^{\circ})}$

$=\frac{2sin^2x+sin^2.cos^260^{\circ}-sin^260^{\circ}.cos^2x}{cos^2x.cos^260^{\circ}-sin^2x.sin^260^{\circ}}$

$=\frac{sin^2x(2+\frac{1}{4})-\frac{3}{4}cos^2x}{\frac{1}{4}cos^2x-\frac{3}{4}sin^2x}$

$=\frac{9sin^2x-3cos^2x}{cos^2x-3sin^2x}=\frac{9-9cos^2x-3cos^2x}{cos^2x-3+3cos^2x}$

$=\frac{9-12cos^2x}{4cos^2x-3}=\frac{-3(4cos^2x-3)}{4cos^2x-3}=-3$

Bài viết đã được chỉnh sửa nội dung bởi lovethislife1997: 20-05-2013 - 12:46