Giải Bài tập 21 trang 107 sgk Toán 11 tập 2 Kết nối

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

$= \frac{1-2sin^{2}x}{cos^{2}2x+sin^{2}2x.sin^{2}x} - \frac{cosx-sinx}{cosx+sinx}$

$=\frac{(cos^2 2x + sin^2 2x.sin^2 x)(1-2sin^2 x)}{(cos^2 2x + sin^2 2x.sin^2 x)}-\frac{(cosx-sinx)^2}{(cosx+sinx)(cosx-sinx)}$

$=\frac{cos^2 2x + sin^2 2x.sin^2 x - 2sin^2 x - cos^2 x - sin^2 x + 2cosx sinx}{cos^2 2x + sin^2 2x.sin^2 x}$

$= \frac{cos^2 2x - cos^2 x + 2sin^2 x - sin^2 x + 2cosx sinx}{cos^2 2x + sin^2 2x.sin^2 x}$ 

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

$= \frac{cos^2 x (2 - 2sin^2 x) + sin^2 x (sin2x)}{cos^2 2x + sin^2 2x.sin^2 x} $

$= \frac{2cos^2 x - 2cos^2 x sin^2 x + sin^2 x sin2x}{cos^2 2x + sin^2 2x.sin^2 x} = \frac{2cos^2 x (1-sin^2 x) + sin^2 x sin2x}{cos^2 2x + sin^2 2x.sin^2 x} $

$= \frac{2cos^2 x cos^2 x + sin^2 x sin2x}{cos^2 2x + sin^2 2x.sin^2 x} $

$= \frac{cos^2 x(2cos^2 x + sin^2 2x)}{cos^2 2x + sin^2 2x.sin^2 x} $

$= \frac{cos^2 x}{1+sin^2 2x}$

b) 

$B = \frac{sin 4x}{1+cos 4x}.\frac{cos2x}{1+cos2x}- cot(\frac{3 \pi}{2} - x)$

$= \frac{2sin2x cos2x}{2cos^2 2x}.\frac{cos2x}{1+cos2x} + tanx $

$= \frac{sin 2x}{cos^2 2x}. \frac{cos2x}{1+cos2x} + tanx $

$= \frac{2sinx cosx}{(1+cos2x)(1-cos2x)}.\frac{1+cos2x}{2} + tanx $

$= \frac{sinx}{cos^2 x} + tanx$

$= \frac{sin^2 x}{sin x cos^2 x} + \frac{sinx}{cosx}$

$= \frac{1}{cos^2 x} + \frac{sinx}{cosx} $

$= sec^2 x + tanx$

c)

$C= 2( cos^{4}x - sin^{4}x)sin2x$

$= 2cos^{4}xsin2x - 2sin^{4}x.sin2x$

$= 2cos^{4}xsin2x - 2sin^{2}x(1-cos^{2}x)sin2x$

$=2cos^{2}x(cos^{2}x-sin^{2}x)sin2x$

$=2cos^{2}x.cos2x.sin2x$

$=4cos^{2}x.cos2xsinx$