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Ta có \(2\sin x\cos x=\left(\sin x+\cos x\right)^2-\left(\sin^2x+\cos^2x\right)\)
\(=\left(\dfrac{3}{4}\right)^2-1=-\dfrac{7}{16}\)
Từ đó \(A=\left|\sin x-\cos x\right|\)
\(\Rightarrow A^2=\left(\sin x-\cos x\right)^2\)
\(A^2=\sin^2x+\cos^2x-2\sin x\cos x\)
\(A^2=1+\dfrac{7}{16}=\dfrac{23}{16}\)
\(\Rightarrow A=\dfrac{\sqrt{23}}{4}\) (do \(A\ge0\))
Có \(\cos x+\sin x=\dfrac{3}{4}\)
\(\Leftrightarrow\left(\cos x+\sin x\right)^2=\dfrac{9}{16}\)
\(\Leftrightarrow2.\sin x.\cos x+1=\dfrac{9}{16}\)
\(\Leftrightarrow\sin x.\cos x=-\dfrac{7}{32}\)
Lại có \(\left(\cos x+\sin x\right)^2=\left(\cos x-\sin x\right)^2+4.\sin x.\cos x=\dfrac{9}{16}\)
\(\Leftrightarrow\left(\cos x-\sin x\right)^2=\dfrac{23}{16}\)
\(\Leftrightarrow\left|\sin x-\cos x\right|=\dfrac{\sqrt{23}}{4}\)
\(M=sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)
\(=sinx.cosx+\dfrac{sin^2x}{\dfrac{cosx+sinx}{sinx}}+\dfrac{cos^2x}{\dfrac{cosx+sinx}{cosx}}\)
\(=sinx.cosx+\dfrac{sin^3x+cos^3x}{cosx+sinx}\)
\(=sinx.cosx+\dfrac{\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{cosx+sinx}\)
\(=sinx.cosx+sin^2x+cos^2x-sinx.cosx\)
\(=sin^2x+cos^2x=1\)
\(\left(sinx+cosx\right)^2=\frac{25}{16}\Rightarrow1+2sinx.cosx=\frac{25}{16}\)
\(\Rightarrow sinx.cosx=\frac{9}{32}\)
\(\left(sinx-cosx\right)^2=\left(sinx+cosx\right)^2-4sinx.cosx=\frac{25}{16}-4.\frac{9}{32}=\frac{7}{16}\)
\(\Rightarrow sinx-cosx=\pm\frac{\sqrt{7}}{4}\)
\(tana-cota=2\sqrt{3}\Rightarrow\left(tana-cota\right)^2=12\)
\(\Rightarrow\left(tana+cota\right)^2-4=12\Rightarrow\left(tana+cota\right)^2=16\)
\(\Rightarrow P=4\)
\(sinx+cosx=\dfrac{1}{5}\Rightarrow\left(sinx+cosx\right)^2=\dfrac{1}{25}\)
\(\Rightarrow1+2sinx.cosx=\dfrac{1}{25}\Rightarrow sinx.cosx=-\dfrac{12}{25}\)
\(P=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}=\dfrac{1}{-\dfrac{12}{25}}=-\dfrac{25}{12}\)
Đáp án C