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

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

\(\Rightarrow P=4\)

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

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

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

\(P=sin^22x-\left[2sin\dfrac{x}{2}cos\dfrac{x}{2}\left(cos^4\dfrac{x}{2}-sin^4\dfrac{x}{2}\right)\right]^2\)

\(=sin^22x-\left[sinx\left(cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}\right)\left(cos^2\dfrac{x}{2}+sin^2\dfrac{x}{2}\right)\right]^2\)

\(=sin^22x-\left[sinx.cosx.1\right]^2\)

\(=sin^22x-\left[\dfrac{1}{2}sin2x\right]^2\)

\(=\dfrac{3}{4}sin^22x=\dfrac{3}{4}\left(1-cos^22x\right)=\dfrac{3}{4}\left(1-\dfrac{1}{4}\right)=\dfrac{9}{16}\)

1.

ĐKXĐ: \(1-x^2>0\Leftrightarrow0< x< 1\)

Pt tương đương:

\(x=5-2m\)

Pt có nghiệm khi và chỉ khi: 

\(0< 5-2m< 1\) \(\Leftrightarrow2< m< \dfrac{5}{2}\)

2.

\(M=\dfrac{\dfrac{sina.cosa}{cos^2a}}{\dfrac{sin^2a}{cos^2a}-\dfrac{cos^2a}{cos^2a}}=\dfrac{tana}{tan^2a-1}=\dfrac{\left(-\dfrac{2}{3}\right)}{\left(-\dfrac{2}{3}\right)^2-1}=-\dfrac{6}{5}\)

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b) \(\sin x+\cos x=\dfrac{3}{2}\)

\(\left(\sin x+\cos x\right)^2=\dfrac{1}{4}\)

\(\sin^2x+\cos^2x+2\sin x\cos x=\dfrac{1}{4}\)

\(2\sin x\cos x=-\dfrac{3}{4}=\sin2x\)

ý a,

ý c

\(B=cos^2x+sin^2x+tan^2x\)

\(=1+tan^2x\)

\(=\dfrac{1}{cos^2x}=1:\dfrac{1}{4}=4\)

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

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

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