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 \(\left(\cos x+\sin x\right)^2=\left(\cos x-\sin x\right)^2+4.\sin x.\cos x=\dfrac{9}{16}\)

\(\Leftrightarrow\left(\cos x-\sin x\right)^2=\dfrac{23}{16}\)

\(\Leftrightarrow\left|\sin x-\cos x\right|=\dfrac{\sqrt{23}}{4}\)

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Đề bị lỗi rồi bạn ơi

\(tana-cota=2\sqrt{3}\Rightarrow\left(tana-cota\right)^2=12\)

\(\Rightarrow\left(tana+cota\right)^2-4=12\Rightarrow\left(tana+cota\right)^2=16\)

\(\Rightarrow P=4\)

\(sinx+cosx=\dfrac{1}{5}\Rightarrow\left(sinx+cosx\right)^2=\dfrac{1}{25}\)

\(\Rightarrow1+2sinx.cosx=\dfrac{1}{25}\Rightarrow sinx.cosx=-\dfrac{12}{25}\)

\(P=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}=\dfrac{1}{-\dfrac{12}{25}}=-\dfrac{25}{12}\)

-4 ở đâu ra vậy ạ

Câu 1: \(P=3\cdot\sin x+4\cdot cosx\)

\(=5\cdot\left(\frac35\cdot\sin x+\frac45\cdot cosx\right)\)

\(=5\cdot\sin\left(x+\alpha\right)\) , với cos α=3/5; sin α=4/5

\(-1\le\sin\left(x+\alpha\right)\le1\)

=>\(-5\le5\sin\left(x+\alpha\right)\le5\)

=>-5<=P<=5

=>P max=5

Câu 2: \(P=\left(\sin^2x-1\right)\cdot\tan^2x+\left(cos^2x-1\right)\cdot\cot^2x\)

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

\(=-cos^2x\cdot\frac{\sin^2x}{cos^2x}-\sin^2x\cdot\frac{cos^2x}{\sin^2x}\)

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