Question

Answer 1

e: tan x=2 nên sin x/cosx=2

=>sinx=2*cosx

\(A=\dfrac{8\cdot cos^3x-2\cdot8\cdot cos^3x}{2cosx-8\cdot cos^3x}=\dfrac{-8\cdot cos^3x}{2\cdot cosx\left(1-4\cdot cos^2x\right)}\)

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

d: \(A=\dfrac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+cot^2x\right)}=\dfrac{sin^2x\cdot\dfrac{1}{cos^2x}}{cos^2x\cdot\dfrac{1}{sin^2x}}\)

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

c: \(VT=sinx\cdot cosx\cdot\left(sin^2x+cos^2x\right)\)

=sinx*cosx*1

=VP

a: Độ dài cung là;

\(C=\dfrac{pi\cdot10\cdot39^027'56''}{180}\simeq6,89\)