đáp án không giống lắm 

Dạ em cảm ơn ạ

a) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{3}{4}=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{cases}}\)

- \(cosx=\frac{1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

- \(cosx=\frac{-1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=-\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\sqrt{3}}=\frac{-\sqrt{3}}{3}\)

b) Bạn làm tương tự câu a) nha. 

1: \(sin^6x+cos^6x+3sin^2x\cdot cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-3\cdot sin^2x\cdot cos^2x\cdot\left(sin^2x+cos^2x\right)+3\cdot sin^2x\cdot cos^2x\)

=1

2: \(sin^4x-cos^4x\)

\(=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\)

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

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

Câu b ko rút gọn được, bạn coi lại đề

\(x^2sin^2a+y^2cos^2a-2xy.sina.cosa+x^2cos^2a+y^2sin^2a+2xy.sinx.cosa\)

\(=x^2\left(sin^2a+cos^2a\right)+y^2\left(cos^2a+sin^2a\right)=x^2+y^2\)

a) \(cos^4x-sin^4x=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=cos^2x-sin^2x\)

b) \(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{tanxcotx}{tanxcotx+cotx}=\frac{1}{1+tanx}+\frac{tanx}{tanx+1}\)

\(=\frac{1+tanx}{1+tanx}=1\)

c) Ta có: \(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}\)

\(\Rightarrow\frac{1}{1+tan^2x}=cos^2x\)

Tương tự \(\frac{1}{1+tan^2y}=cos^2y\)

\(\Rightarrow cos^2x-cos^2y=\frac{1}{1+tan^2x}-\frac{1}{1+tan^2y}\)

\(cos^2x-cos^2y=\left(1-sin^2x\right)-\left(1-sin^2y\right)=sin^2y-sin^2x\)

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

\(A=s\left(x\right)cs\left(x\right)+\frac{\left(s^3\left(x\right)+cs^3\left(x\right)\right)}{cs\left(x\right)\left(1+t\left(x\right)\right)}=s\left(x\right)cs\left(x\right)+\left(\frac{\left(s\left(x\right)+cs\left(x\right)\right)\left(1-s\left(x\right)cs\left(x\right)\right)}{\left(s\left(x\right)+cs\left(x\right)\right)}\right)\)

\(=1\) vì \(s\left(x\right)+cs\left(x\right)\ne0,\forall0< =x< =\frac{\pi}{2}\)

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

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

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

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