Question

[nguyễn duy khánh]

chứng minh 
a> cot²x/1+cot²x   .  1+tan²x/tan²x  = tan²x+cot²x/1=tan⁴x
b>tan²x-cos²x/sin²x    +    cot²x-sin²x/cos²x  = 2
 

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(VT=\dfrac{cot^2x}{1+cot^2x}\cdot\dfrac{1+tan^2x}{tan^2x}\)

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

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

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

\(=cot^2x\)

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

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

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

=>VT=VP

b:

\(\dfrac{tan^2x-cos^2x}{sin^2x}+\dfrac{cot^2x-sin^2x}{cos^2x}\)

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

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

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

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

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