\(a,1+tan^2x=\dfrac{1}{cos^2x}\\ VT=1+\dfrac{sin^2x}{cos^2x}\\ =\dfrac{cos^2x}{cos^2x}+\dfrac{sin^2x}{cos^2x}\\ =\dfrac{sin^2x+cos^2x}{cos^2x}=\dfrac{1}{cos^2x}=VP\)

\(b,VT=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}\\ =\dfrac{sin^2x+cos^2x}{cosx.sinx}=\dfrac{1}{cosx.sinx}=VP\)

a: tan x(cot^2x-1)

\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)

=cotx-tanx/cotx=cotx(1-tan^2x)

b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)

\(=sin^2x\left(\dfrac{1}{cos^2x}-1\right)=sin^2x\cdot\dfrac{sin^2x}{cos^2x}=sin^2x\cdot tan^2x\)

c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)

\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)

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

=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)

=-cos^2x*cos^2x=-cos^4x

=>ĐPCM

Đề bài sai

Ví dụ: với \(a=1;b=2;c=3,d=4\) thì \(x=\dfrac{1}{2}\) ; \(y=\dfrac{3}{4}\) ; \(z=\dfrac{2}{3}\)

Khi đó  \(x< y\) nhưng \(z< y\)

\(\text{Vì }\dfrac{a}{b}< \dfrac{c}{d}\text{ nên }ad< bc\left(1\right)\)

\(\text{Xét tích}:a\left(b+d\right)=ab+ad\left(2\right)\)

                \(b\left(a+c\right)=ba+bc\left(3\right)\)

\(\text{Từ(1);(2);(3)}\Rightarrow a\left(b+d\right)< b\left(a+c\right)\text{ do đó }\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(4\right)\)

\(\text{Tương tự ta có:}\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(5\right)\)

\(\text{Từ (4);(5) ta được }\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)

\(\Rightarrow x< y< z\)

Đúng như bạn viết vế trái là thế này:

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

\(=\left(\frac{1}{cot^2x+1}\right)\left(\frac{1+cot^2x}{cotx}\right)=\frac{1}{cotx}=tanx\)

Còn vế phải sẽ ra thế này:

\(\frac{1+tan^4x}{tan^2x+cot^2x}=\frac{1+tan^4x}{tan^2x+\frac{1}{tan^2x}}=\frac{tan^2x\left(1+tan^4x\right)}{tan^4x+1}=tan^2x\)

Hai vế ra kết quả khác nhau nên chắc bạn ghi sai đề :)

Yêu cầu của đề là gì vậy bạn?

Ta có:

\(\left(x+y+z\right)\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z\)

\(\Leftrightarrow x+y+z=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z\)

\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0\)

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