1) Chứng minh rằng:

a) $\frac{1+\sin 2x+\cos 2x}{1+\sin 2x-\cos 2x}=\cot x$

b) $\frac{{\cos }^2\alpha -{\sin }^2\alpha }{{\cot }^2\alpha -{\tan }^2\alpha }={\sin }^2\alpha {\cos }^2\alpha$

c) $\frac{\tan \alpha -\tan \beta }{\cot \beta -\cot \alpha }=\tan \alpha \tan \beta$

d) $\frac{1-4{\sin }^{2}x{\cos }^{2}x}{(\sin x+\cos x{)}^2}=(\sin x-\cos x{)}^2$

2) Cho tam giác $ABC$, chứng minh rằng $\sin (A+B+2C)=-\sin C$.

(AI giải đề cho tôi)
Bài 2 (Phần 4 — Tự luận), lời giải đầy đủ:

1a) VT $=\frac{1+\sin 2x+\cos 2x}{1+\sin 2x-\cos 2x}=\frac{(\sin x+\cos x{)}^2+\cos 2x}{(\sin x+\cos x{)}^2-\cos 2x}$. Ta có $1+\sin 2x=(\sin x+\cos x{)}^2$, nên:
Tử số $=(\sin x+\cos x{)}^2+(\cos x+\sin x)(\cos x-\sin x)=(\sin x+\cos x)[(\sin x+\cos x)+(\cos x-\sin x)]=2\cos x(\sin x+\cos x)$.
Mẫu số $=(\sin x+\cos x{)}^2-(\cos x+\sin x)(\cos x-\sin x)=(\sin x+\cos x)[(\sin x+\cos x)-(\cos x-\sin x)]=2\sin x(\sin x+\cos x)$.
Vậy VT $=\frac{2\cos x(\sin x+\cos x)}{2\sin x(\sin x+\cos x)}=\cot x$ (điều kiện $\sin x\ne 0$, $\sin x+\cos x\ne 0$). Đpcm.

1b) VT $=\frac{{\cos }^2\alpha -{\sin }^2\alpha }{\frac{{\cos }^2\alpha }{{\sin }^2\alpha }-\frac{{\sin }^2\alpha }{{\cos }^2\alpha }}=\frac{({\cos }^2\alpha -{\sin }^2\alpha ){\sin }^2\alpha {\cos }^2\alpha }{{\cos }^4\alpha -{\sin }^4\alpha }=\frac{({\cos }^2\alpha -{\sin }^2\alpha ){\sin }^2\alpha {\cos }^2\alpha }{({\cos }^2\alpha -{\sin }^2\alpha )({\cos }^2\alpha +{\sin }^2\alpha )}={\sin }^2\alpha {\cos }^2\alpha =$ VP (với $\sin \alpha \cos \alpha \ne 0$, ${\cos }^2\alpha \ne {\sin }^2\alpha$). Đpcm.

1c) VT $=\frac{\tan \alpha -\tan \beta }{\frac{1}{\tan \beta }-\frac{1}{\tan \alpha }}=\frac{\tan \alpha -\tan \beta }{\frac{\tan \alpha -\tan \beta }{\tan \alpha \tan \beta }}=\tan \alpha \tan \beta =$ VP (với $\tan \alpha \tan \beta \ne 0$, $\tan \alpha \ne \tan \beta$). Đpcm.

1d) VT $=\frac{1-4{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x+2\sin x\cos x+{\cos }^{2}x}=\frac{({\sin }^{2}x+{\cos }^{2}x{)}^2-4{\sin }^{2}x{\cos }^{2}x}{1+2\sin x\cos x}=\frac{({\sin }^{2}x-{\cos }^{2}x{)}^2}{1+2\sin x\cos x}=\frac{(\sin x-\cos x{)}^2(\sin x+\cos x{)}^2}{(\sin x+\cos x{)}^2}=(\sin x-\cos x{)}^2=$ VP (với $\sin x+\cos x\ne 0$). Đpcm.

2) Vì $A,B,C$ là ba góc của tam giác nên $A+B+C=\pi$, suy ra $A+B=\pi -C$. Do đó:
$\sin (A+B+2C)=\sin [(A+B+C)+C]=\sin (\pi +C)=-\sin C$. Đpcm.