Đáp án:

1)

Ta có:

$(1+\cos{\alpha})(1+\tan{\alpha})\\=1+\tan{\alpha}+\cos{\alpha}+\cos{\alpha}\tan{\alpha}\\=1+\tan{\alpha}+\cos{\alpha}+\cos{\alpha}.\dfrac{\sin{\alpha}}{\cos{\alpha}}\,\,\,\left(\tan{\alpha}=\dfrac{\sin{\alpha}}{\cos{\alpha}}\right)\\=1+\tan{\alpha}+\cos{\alpha}+\sin{\alpha}$

Vậy $1+\sin{\alpha}+\cos{\alpha}+\tan{\alpha}=(1+\cos{\alpha})(1+\tan{\alpha})$

2)

Ta có:

$\tan{\alpha}+\cot{\alpha}\\=\dfrac{\sin{\alpha}}{\cos{\alpha}}+\dfrac{\cos{\alpha}}{\sin{\alpha}}\,\,\,\left(\tan{\alpha}=\dfrac{\sin{\alpha}}{\cos{\alpha}};\cot{\alpha}=\dfrac{\cos{\alpha}}{\sin{\alpha}}\right)\\=\dfrac{\sin^2{\alpha}+\cos^2{\alpha}}{\sin{\alpha}\cos{\alpha}}\\=\dfrac{1}{\sin{\alpha}\cos{\alpha}}\,\,\,(\sin^2{\alpha}+\cos^2{\alpha}=1)\\=\dfrac{2}{2\sin{\alpha}\cos{\alpha}}\\=\dfrac{2}{\sin{2\alpha}}\,\,\,(2\sin{\alpha}\cos{\alpha}=\sin{2\alpha})$

Vậy $\tan{\alpha}+\cot{\alpha}=\dfrac{2}{\sin{2\alpha}}$

3)

Ta có:

$\cot{\alpha}-\tan{\alpha}\\=\dfrac{\cos{\alpha}}{\sin{\alpha}}-\dfrac{\sin{\alpha}}{\cos{\alpha}}\\=\dfrac{\cos^2{\alpha}-\sin^2{\alpha}}{\sin{\alpha}\cos{\alpha}}\\=\dfrac{\cos{2\alpha}}{\sin{\alpha}\cos{\alpha}}\,\,\,(\cos^2{\alpha}-\sin^2{\alpha}=\cos{2\alpha})\\=\dfrac{2\cos{2\alpha}}{2\sin{\alpha}\cos{\alpha}}\\=\dfrac{2\cos{2\alpha}}{\sin{2\alpha}}\,\,\,(2\sin{\alpha}\cos{\alpha}=\sin{2\alpha})\\=2\cot{2\alpha}$

Vậy $\cot{\alpha}-\tan{\alpha}=2\cot{2\alpha}$

Giải thích các bước giải:

Sử dụng:

+ Công thức nhân đôi: $\sin{2\alpha}=2\sin{\alpha}\cos{\alpha};\cos{2\alpha}=\cos^2{\alpha}-\sin^2{\alpha}$

+ Công thức $\tan{\alpha}=\dfrac{\sin{\alpha}}{\cos{\alpha}};\cot{\alpha}=\dfrac{\cos{\alpha}}{\sin{\alpha}}$