Tính giá trị biểu thức

Question

$A=tan110^{o}.tan340^{o}+sin160^{o}.cos110^{o}+sin250^{o}.cos340^{o}$

$B=\frac{tan225^{o}-cot81^{o}.cot69^{o}}{cot261^{o}+tan201^{o}}$

score 3 · Answer 1

$A=tan110^0.tan340^0+sin160^0.cos110^0+sin250^0.cos340^0$

$=tan(90^0+20^0).tan(360^0-20^0)+sin(180^0-20^0).cos(90^0+20^0)+sin[360^0-(90^0+20^0)].cos(360^0-20^0)$

$=cos20^0.tan20^0-sin^220^0-cos^220^0=cos^220^0-cos^220^0=0$

$B=\frac{tan225^0-cot81^0.cot69^0}{cot261^0+tan201^0}$

$=\frac{tan(180^0+45^0)-cot(90^0-9^0).cot(90^0-21^0)}{cot(180^0+90^0-9^0)+tan(180^0+21^0)}$

$=\frac{1-tan9^0.tan21^0}{tan9^0+tan21^0}=\frac{1}{tan(9^0+21^0)}=\sqrt{3}$

Thấy đúng thì nhấn vào biểu tượng chữ V màu trắng bên trái đáp án nhé! Lần sau mình sẽ ss giúp đỡ. Tks bạn! – Nero 09-05-14 07:19 PM

score 2 · Answer 2

$A=tan110^{o}.tan340^{o}+sin160^{o}.cos110^{o}+sin250^{o}.cos340^{o}$

$=tan(90^{o}+20^{o}).tan(360^{o}-20^{o})+sin(180^{o}-20^{o}).cos(90^{o}+20^{o})+sin(160^{o}+90^{o})cos(160^{o}+180^{o})$

$=cot20^{o}.tan20^{o}-sin20^{o}.sin20^{o}-cos160^{o}.cos160^{o}$

$=1-[sin^{2}20^{o}+cos^{2}(180^{o}-20^{o})]$

$=1-(sin^{2}20^{o}+cos^{2}20^{o})=1-1=0$

2 Đáp án