tính A = $\frac{2\cos x}{1-\sin x}$ 

biết tan $\frac{x}{2}$  = $\frac{4}{5}$
ban xem lai dap an di hinh nhu sai doban ay ap dung cong thuc sai o mau –  minhquangtranquangkhai 26-04-14 09:22 PM
$A=\frac{2(cos^2\frac x2-sin^2\frac x2)}{1-2sin\frac x2cos\frac x2}=\frac{2(cos\frac x2-sin\frac x2)(cos\frac x2+sin\frac x2)}{(cos\frac x2-sin\frac x2)^2}$
$=\frac{2cos\frac x2+2sin\frac x2}{cos\frac x2-sin\frac x2}=\frac{2+2tan\frac x2}{1-tan\frac x2}=18$
Ta có: $A=\frac{2(cos^2\frac{x}{2}-sin^2\frac{x}{2})}{(cos\frac{x}{2}-sin\frac{x}{2})^2}=\frac{2(cos\frac{x}{2}+sin\frac{x}{2})}{cos\frac{x}{2}-sin\frac{x}{2}}$
Chia cả tử và mẫu cho $cos\frac{x}{2}$
$\Rightarrow A=\frac{2+2tan\frac{x}{2}}{1-tan\frac{x}{2}}=18$

ukm,tks bạn nhé.k sao mà –  girlkute_9x_nd98 26-04-14 09:36 PM
bạn thông cảm nhé! mình nhìn nhầm sinx thành cosx. Mình thật sự xin lỗi bạn về sự cố nhé :D –  Nero 26-04-14 09:34 PM
nhầm rồi nhé bạn! –  Nero 26-04-14 09:26 PM
Thấy đúng thì nhấn vào biểu tượng chữ V bên dưới vote down nhé! Lần sau mình sẽ ss giúp đỡ. Tks bạn! :D –  Nero 26-04-14 09:11 PM

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