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Sin 1 slot xíu nữa làm

\(1-\frac{\sin^2x}{1+\cot x}-\frac{\cos^2x}{1+\tan x}\)

\(=1\left(\frac{\sin^2x}{1+\frac{\cos x}{\sin x}}+\frac{\cos^2x}{1+\frac{\sin x}{\cos x}}\right)\)

\(=1-\left(\frac{\sin^2x}{\frac{\sin x+\cos x}{\sin x}}+\frac{\cos^2x}{\frac{\cos x+\sin x}{\cos x}}\right)\)

\(=1-\left(\frac{\sin^3x}{\sin x+\cos x}+\frac{\cos^3x}{\sin x+\cos x}\right)\)

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

\(=1-\)\(\frac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cos x+\cos^2x\right)}{\sin x+\cos x}\)

\(=\sin x\cos x\)

bạn chỉ cần nhớ rằng: sin²x+ cos²x= 1 và cotx*tanx= 1 rồi quy đồng lên và làm bình thường

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

\(=\left(1+\dfrac{cosx}{sinx}-sin^2x\right):\dfrac{sinx+cosx}{sinx}-cos^2x:\dfrac{cosx+sinx}{cosx}\)

\(=\dfrac{sinx+cosx-sin^3x}{sinx}\cdot\dfrac{sinx}{sinx+cosx}-\dfrac{cos^3x}{cosx+sinx}\)

\(=\dfrac{sinx+cosx-sin^3x-cos^3x}{sinx+cosx}\)

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

\(=1-1+sinx\cdot cosx=\dfrac{1}{2}sin2x\)

Chọn C.

Ta có:

\(M=sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)

\(=sinx.cosx+\dfrac{sin^2x}{\dfrac{cosx+sinx}{sinx}}+\dfrac{cos^2x}{\dfrac{cosx+sinx}{cosx}}\)

\(=sinx.cosx+\dfrac{sin^3x+cos^3x}{cosx+sinx}\)

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

\(=sinx.cosx+sin^2x+cos^2x-sinx.cosx\)

\(=sin^2x+cos^2x=1\)

Chọn B.

Ta có: A = (tanx + cotx)² - ( tanx - cotx)²

= tan²x +  2tanx.cot x + cot²x - ( tan²x - 2tanx.cotx + cot²x)

= 4tanx.cotx = 4.

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

\(=\frac{\sin^3x+\cos^3x}{\sin x+\cos x}-1=\frac{\left(\sin x+\cos x\right).\left(\sin^2x-\sin x.\cos x+\cos^2x\right)}{\sin x+\cos x}-1\)

\(=1-\sin x.\cos x-1=-\sin x.\cos x\)