Bài 21 trang 107 SGK Toán 11 tập 2 - Kết nối tri thức

Đề bài

Rút gọn các biểu thức sau:

a) \(A = \frac{{1 - 2{{\sin }^2}x}}{{1 + \sin 2x}} - \frac{{1 - \tan x}}{{1 + \tan x}}\)

b) \(B = \frac{{\sin 4x}}{{1 + \cos 4x}} \cdot \frac{{\cos 2x}}{{1 + \cos 2x}} - \cot \left( {\frac{{3\pi }}{2} - x} \right)\);

c) \(C = 2\left( {{{\cos }^4}x - {{\sin }^4}x} \right)\sin 2x\).

Lời giải chi tiết

a) \(A = \dfrac{{1 - 2{{\sin }^2}x}}{{1 + \sin 2x}} - \dfrac{{1 - \tan x}}{{1 + \tan x}} \)

\(= \dfrac{{{{\sin }^2}x + {{\cos }^2}x - 2{{\sin }^2}x}}{{{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x}} - \dfrac{{1 - \dfrac{{\sin x}}{{\cos x}}}}{{1 + \dfrac{{\sin x}}{{\cos x}}}}\)

\(= \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\sin x + \cos x} \right)}^2}}} - \dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}\)

\(= \dfrac{{\cos x - \sin x}}{{\sin x + \cos x}} - \dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} = 0\).

b) \(B = \dfrac{{\sin 4x}}{{1 + \cos 4x}} \cdot \dfrac{{\cos 2x}}{{1 + \cos 2x}} - \cot \left( {\dfrac{{3\pi }}{2} - x} \right)\)

\(= \dfrac{{2\sin 2x\cos 2x}}{{2{{\cos }^2}2x}}.\dfrac{{\cos 2x}}{{2{{\cos }^2}x}} - \cot \left( {\dfrac{\pi }{2} - x} \right)\)

\(= \dfrac{{\sin 2x}}{{2{{\cos }^2}x}} - \tan x\)

\(= \dfrac{{\sin 2x - 2\sin x\cos x}}{{2{{\cos }^2}x}} = 0\).

c) \(C = 2\left( {{{\cos }^4}x - {{\sin }^4}x} \right)\sin 2x\)

\(= 2\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right)\sin 2x\)

\(= 2\cos 2x.\sin 2x = \sin 4x\).