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\(\frac{2}{sin4x}-tan2x=\frac{2}{2sin2x.cos2x}-\frac{sin2x}{cos2x}=\frac{1}{cos2x}\left(\frac{1}{sin2x}-sin2x\right)\)
\(=\frac{1}{cos2x}\left(\frac{1-sin^22x}{sin2x}\right)=\frac{1}{cos2x}\frac{cos^22x}{sin2x}=\frac{cos2x}{sin2x}=cot2x\)
\(\frac{sin2x-sin4x}{1-cos2x+cos4x}=\frac{sin2x-2sin2x.cos2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(1-2cos2x\right)}{-cos2x\left(1-2cos2x\right)}=\frac{-sin2x}{cos2x}=-tan2x\)
\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=-\left(\frac{sin2x-sin4x}{1-cos2x+cos4x}\right)=-\left(-tan2x\right)=tan2x\) lấy luôn kết quả câu trên cho lẹ, biến đổi thì làm y hệt
a: Ta sẽ có hình vẽ sau:
Đặt \(x=\widehat{B}\)
sin x=sin B=AC/BC
cosx=cosB=AB/BC
\(tanx=tanB=\dfrac{AC}{AB}=\dfrac{sinx}{cosx}\)
=>\(tan^2x=\dfrac{sin^2x}{cos^2x}\)
b: \(cot^2x=\dfrac{1}{tan^2x}=1:\dfrac{sin^2x}{cos^2x}=\dfrac{cos^2x}{sin^2x}\)
Đáp án: C
Ta có:
A = (1 - sin 2 x ) c o t 2 x + (1 - c o t 2 x ) = c o t 2 x - sin 2 x . c o t 2 x + 1 - c o t 2 x
Chọn A.
Ta có C = (1-sin2x) cot2x + 1 - cot2x.
= (1 - sin2x - 1) cot2x + 1
= -sin2x.cot2x + 1 = -cos2x + 1 = sin2x.
`B=(sin2x)/(tanx+cot2x)`
Tử ` = 2sinxcosx`
Mẫu `=(sinx)/(cosx) + (cos2x)/(sin2x)`
`=(sinx . sin2x + cosx .cos2x)/(2sinx cosx . cosx)`
`=(cos (2x-x))/(2sinxcos^2x)`
`=(cosx)/(2sinxcos^2x)`
`=1/(2sinxcosx)`
`=> B = sin^2 2x`
Lớp 8 nên không chắc ạ.
\(B=\dfrac{sin2x}{tanx+cot2x}=\dfrac{2sinx.cosx}{\dfrac{sinx}{cosx}+\dfrac{cos2x}{sin2x}}=\dfrac{2sinx.cosx}{\dfrac{sinx.sin2x+cos2x.cosx}{cosx.sin2x}}=\dfrac{2sinx.cosx}{\dfrac{.2sin^2x.cosx+cosx\left(2cos^2x-1\right)}{cosx.2sinx.cosx}}=\dfrac{2sinx.cosx.}{\dfrac{cosx\left(2sin^2x+2cos^2x-1\right)}{cos.2sinx.cosx}}=\dfrac{2sinx.cosx}{\dfrac{1}{2sinx.cosx}}=2sinx.cosx.2sinx.cosx=sin^22x.\)
\(=\left(\dfrac{2sinx.cosx}{cos2x}-\dfrac{sinx}{cosx}\right)\left(2sinx.cosx-\dfrac{sinx}{cosx}\right)\)
\(=sinx\left(\dfrac{2cosx}{cos2x}-\dfrac{1}{cosx}\right).sinx\left(2cosx-\dfrac{1}{cosx}\right)\)
\(=sin^2x\left(\dfrac{2cos^2x-\left(2cos^2x-1\right)}{cosx.cos2x}\right)\left(\dfrac{2cos^2x-1}{cosx}\right)\)
\(=sin^2x\left(\dfrac{1}{cosx.cos2x}\right)\left(\dfrac{cos2x}{cosx}\right)=\dfrac{sin^2x}{cos^2x}=tan^2x\)
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:
\(\dfrac{tan^2x-cos^2x}{sin^2x}+\dfrac{cot^2x-sin^2x}{cos^2x}\)
\(=\dfrac{\left(\dfrac{sinx}{cosx}\right)^2-cos^2x}{sin^2x}+\dfrac{\left(\dfrac{cosx}{sinx}\right)^2-sin^2x}{cos^2x}\)
\(=\dfrac{sin^2x-cos^4x}{cos^2x\cdot sin^2x}+\dfrac{cos^2x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x+cos^2x-cos^4x-sin^4x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{1-\left(cos^2x+sin^2x\right)^2+2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}=2\)