\(A+B+C=180^0\Rightarrow\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^0\Rightarrow\frac{A}{2}+\frac{B}{2}=90^0-\frac{C}{2}\)

\(\Rightarrow tan\left(\frac{A}{2}+\frac{B}{2}\right)=tan\left(90^0-\frac{C}{2}\right)\)

\(\Leftrightarrow\frac{tan\frac{A}{2}+tan\frac{B}{2}}{1-tan\frac{A}{2}.tan\frac{B}{2}}=cot\frac{C}{2}=\frac{1}{tan\frac{C}{2}}\)

\(\Leftrightarrow tan\frac{C}{2}\left(tan\frac{A}{2}+tan\frac{B}{2}\right)=1-tan\frac{A}{2}.tan\frac{B}{2}\)

\(\Leftrightarrow tan\frac{A}{2}tan\frac{C}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{A}{2}.tan\frac{B}{2}=1\)

b/\(A+B+C=180^0\Rightarrow A+B=180^0-C\)

\(\Rightarrow cot\left(A+B\right)=cot\left(180^0-C\right)\)

\(\Leftrightarrow\frac{cotA.cotB-1}{cotA+cotB}=-cotC\)

\(\Leftrightarrow cotA.cotB-1=-cotA.cotC-cotB.cotC\)

\(\Leftrightarrow cotA.cotB+cotB.cotC+cotA.cotC=1\)

\(A=tan\left(a+b\right)=tan\frac{\pi}{4}=1\)

Ta có: \(tan\left(a+b\right)=\frac{tana+tanb}{1-tana.tanb}\)

\(\Rightarrow B=tana+tanb=tan\left(a+b\right)\left(1-tana.tanb\right)=1.\left(1-3+2\sqrt{2}\right)=2\sqrt{2}-2\)

\(\left\{{}\begin{matrix}tana+tanb=2\sqrt{2}-2\\tana.tanb=3-2\sqrt{2}\end{matrix}\right.\)

Theo Viet đảo, \(tana;tanb\) là nghiệm của:

\(x^2-\left(2\sqrt{2}-2\right)x+3-2\sqrt{2}=0\)

\(\Leftrightarrow\left(x-\sqrt{2}+1\right)^2=0\Rightarrow x=\sqrt{2}-1\)

\(\Rightarrow tana=tanb=\sqrt{2}-1\Rightarrow a=b=\frac{\pi}{8}\)

\(tan^2a+cot^2a=\left(tana+cota\right)^2-2=m^2-2\)

\(tan^4a+cot^4a=\left(tan^2a+cot^2a\right)^2-2=\left(m^2-2\right)^2-2\)

\(tan^6a+cot^6a=\left(tan^2a+cot^2a\right)^3-3\left(tan^2a+cot^2a\right)\)

\(=\left(m^2-2\right)^3-3\left(m^2-2\right)\)

\(m^2=\left(tana+cota\right)^2=\left(tana-cota\right)^2+4tana.cota\)

\(\Rightarrow m^2=\left(tana-cota\right)^2+4\ge4\)

\(\Rightarrow\left|m\right|\ge2\)

Câu a)

Ta sử dụng 2 công thức:

\(\bullet \tan (180-\alpha)=-\tan \alpha\)

\(\bullet \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha.\tan \beta}\)

Áp dụng vào bài toán:

\(\text{VT}=\tan A+\tan B+\tan C=\tan A+\tan B+\tan (180-A-B)\)

\(=\tan A+\tan B-\tan (A+B)=\tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A.\tan B}\)

\(=(\tan A+\tan B)\left(1+\frac{1}{1-\tan A.\tan B}\right)=(\tan A+\tan B).\frac{-\tan A.\tan B}{1-\tan A.\tan B}\)

\(=-\tan A.\tan B.\frac{\tan A+\tan B}{1-\tan A.\tan B}=-\tan A.\tan B.\tan (A+B)\)

\(=\tan A.\tan B.\tan (180-A-B)\)

\(=\tan A.\tan B.\tan C=\text{VP}\)

Do đó ta có đpcm

Tam giác $ABC$ có ba góc nhọn nên \(\tan A, \tan B, \tan C>0\)

Áp dụng BĐT Cauchy ta có:

\(P=\tan A+\tan B+\tan C\geq 3\sqrt[3]{\tan A.\tan B.\tan C}\)

\(\Leftrightarrow P=\tan A+\tan B+\tan C\geq 3\sqrt[3]{\tan A+\tan B+\tan C}\)

\(\Rightarrow P\geq 3\sqrt[3]{P}\)

\(\Rightarrow P^3\geq 27P\Leftrightarrow P(P^2-27)\geq 0\)

\(\Rightarrow P^2-27\geq 0\Rightarrow P\geq 3\sqrt{3}\)

Vậy \(P_{\min}=3\sqrt{3}\). Dấu bằng xảy ra khi \(\angle A=\angle B=\angle C=60^0\)

Câu b)

Ta sử dụng 2 công thức chính:

\(\bullet \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha.\tan \beta}\)

\(\bullet \tan (90-\alpha)=\frac{1}{\tan \alpha}\)

Áp dụng vào bài toán:

\(\text{VT}=\tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{B}{2}.\tan \frac{C}{2}+\tan \frac{C}{2}.\tan \frac{A}{2}\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{C}{2}(\tan \frac{A}{2}+\tan \frac{B}{2})\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\tan (90-\frac{A+B}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{\tan (\frac{A+B}{2})}\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}.\tan \frac{B}{2}}}\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+1-\tan \frac{A}{2}.\tan \frac{B}{2}=1=\text{VP}\)

Ta có đpcm.

Cũng giống phần a, ta biết do ABC là tam giác nhọn nên

\(\tan A, \tan B, \tan C>0\)

Đặt \(\tan A=x, \tan B=y, \tan C=z\). Ta có: \(xy+yz+xz=1\)

Và \(T=x+y+z\)

\(\Rightarrow T^2=x^2+y^2+z^2+2(xy+yz+xz)\)

Theo hệ quả quen thuộc của BĐT Cauchy:

\(x^2+y^2+z^2\geq xy+yz+xz\)

\(\Rightarrow T^2\geq 3(xy+yz+xz)=3\)

\(\Rightarrow T\geq \sqrt{3}\Leftrightarrow T_{\min}=\sqrt{3}\)

Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\Leftrightarrow \angle A=\angle B=\angle C=60^0\)

\(cos^4a+sin^4a-6sin^2a.cos^2a\)

\(=cos^4a+sin^4a-2sin^2a.cos^2a-4sin^2a.cos^2a\)

\(=\left(cos^2a-sin^2a\right)^2-\left(2sina.cosa\right)^2\)

\(=cos^22a-sin^22a\)

\(=cos4a\)

a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).

b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).