Question

Câu 40: Cho góc nhọn có số đo \(x=\dfrac{1}{2}\) và \(F=tan^2x-sin^2x.tan^2x\). Giá trị của \(F\) bằng?

Answer 1

\(F=tan^2x\left(1-sin^2x\right)=tan^2x\cdot cos^2x\)

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

\(F=sin^2\left(\dfrac{1}{2}\right)\simeq7,62\cdot10^{-5}\)

Answer 2

`F = tan^2x ( 1 - sin^2x ) = tan^2x . cos^2x = ( sin^2x ) / ( cos^2x) . cos^2x = sin^2x`

Thay `x = 1/2,` ta có :

`F = sin^2x . 1/2 ≃ 76,2 . 10^(-5)`