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:

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

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

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

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

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

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

Nhân cả tử và mẫu vế trái với \(cos2x.cosx\) ta được:

\(\frac{sin2x.sinx}{sin2x.cosx-cos2x.sinx}=\frac{sin2x.sinx}{sin\left(2x-x\right)}=\frac{sin2x.sinx}{sinx}=sin2x\)

\(=\left(\dfrac{2sinx.cosx}{cos2x}-\dfrac{sinx}{cosx}\right)\left(2sinx.cosx-\dfrac{sinx}{cosx}\right)\)

\(=sinx\left(\dfrac{2cosx}{cos2x}-\dfrac{1}{cosx}\right).sinx\left(2cosx-\dfrac{1}{cosx}\right)\)

\(=sin^2x\left(\dfrac{2cos^2x-\left(2cos^2x-1\right)}{cosx.cos2x}\right)\left(\dfrac{2cos^2x-1}{cosx}\right)\)

\(=sin^2x\left(\dfrac{1}{cosx.cos2x}\right)\left(\dfrac{cos2x}{cosx}\right)=\dfrac{sin^2x}{cos^2x}=tan^2x\)

\(\frac{sin2x-sin4x}{1-cos2x+cos4x}=\frac{sin2x-2sin2x.cos2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(1-2cos2x\right)}{-cos2x\left(1-2cos2x\right)}=\frac{-sin2x}{cos2x}=-tan2x\)

\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=-\left(\frac{sin2x-sin4x}{1-cos2x+cos4x}\right)=-\left(-tan2x\right)=tan2x\) lấy luôn kết quả câu trên cho lẹ, biến đổi thì làm y hệt

ta có : \(\dfrac{sin2x}{tan\left(\dfrac{\pi}{4}-x\right)\left(1+sin2x\right)}=\dfrac{sin2x}{tan\left(-\left(x-\dfrac{\pi}{4}\right)\right)\left(sin^2x+2sinx.cosx+cos^2x\right)}\)

\(=\dfrac{sin2x}{-tan\left(x-\dfrac{\pi}{4}\right)\left(sinx+cosx\right)^2}=\dfrac{sin2x}{-\dfrac{sin\left(x-\dfrac{\pi}{4}\right)}{cos\left(x-\dfrac{\pi}{4}\right)}\left(sinx+cosx\right)^2}\)

\(=\dfrac{sin2x}{-\dfrac{\dfrac{sinx-cosx}{\sqrt{2}}}{\dfrac{sinx+cosx}{\sqrt{2}}}\left(sinx+cosx\right)^2}=\dfrac{sin2x}{-\left(\dfrac{sinx-cosx}{sinx+cosx}\right)\left(sinx+cosx\right)^2}\)

\(=\dfrac{sin2x}{-\left(sinx-cosx\right)\left(sinx+cosx\right)}=\dfrac{sin2x}{-\left(sin^2x-cos^2x\right)}\)

\(=\dfrac{sin2x}{cos^2x-sin^2x}=\dfrac{sin2x}{cos2x}=tan2x\left(đpcm\right)\)

\(180^o< x< 270^o\)

\(1+tan^2x=\dfrac{1}{cos^2x}\Leftrightarrow1+3^2=\dfrac{1}{cos^2x}\Leftrightarrow cosx=-\dfrac{\sqrt{10}}{10}\)

\(sinx=tanx.cosx=-\dfrac{3\sqrt{10}}{10}\)

\(sin2x=2sinxcosx=2.\left(-\dfrac{3\sqrt{10}}{10}\right).\left(-\dfrac{\sqrt{10}}{10}\right)=\dfrac{3}{5}\)

\(tan2x=\dfrac{2tanx}{1-tan^2x}=\dfrac{2.3}{1-3^2}=-\dfrac{3}{4}\)

\(cos4x=8cos^4x-8cos^2x+1=8.\left(-\dfrac{\sqrt{10}}{10}\right)^4-8.\left(-\dfrac{\sqrt{10}}{10}\right)^2+1=\dfrac{7}{25}\)

tới đây rồi làm tiếp nhé, bận=)

Xét tam giác ABC vuông tại A có AH là đường cao và AM là trung tuyến

Đặt \(\widehat{MAC}=\widehat{MCA}=x\)thì \(\widehat{BMA}=2x\)(theo tính chất đường trung tuyến ứng với cạnh huyền của tam giác vuông)

a) Ta có: \(\sin2x=\frac{AH}{AM}=2.\frac{AH}{BC}=2.\frac{AH}{AC}.\frac{AC}{BC}=2.\sin ACH.\cos ACB=2\cos x.\sin x\)

b) \(\cos2x=\frac{HM}{AM}=\frac{2HM}{BC}=\frac{2HC-2CM}{BC}=2.\frac{HC}{BC}-1=2.\frac{HC}{ AC}.\frac{AC}{BC}-1=2.\cos ACH.\cos ACB-1=2\cos^2x-1=2\cos^2x-\left(\sin^2x+\cos^2x\right)=\cos^2x-\sin^2x\)c) \(\tan2x=\frac{\sin2x}{\cos2x}=\frac{2\cos x.\sin x}{\cos^2x-\sin^2x}=\frac{2.\frac{\sin x}{\cos x}}{\frac{\cos^2x}{\cos^2x}-\frac{\sin^2x}{\cos^2x}}=\frac{2\tan x}{1-\tan^2x}\)