Ta có \(A+B+C=\pi\)

\(\Rightarrow A+B=\pi-C\)

\(\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)

\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\)

\(\Rightarrow tanA+tanB=-tanC\left(1-tanA.tanB\right)\)

\(\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)

\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\) ( đpcm )

a)\(VT=sinA+sinB+sinC=2sin\frac{A+B}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)

\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)(đpcm)

b)Ta có:\(A+B+C=180^O\)

\(\Rightarrow tan\left(A+B\right)=tan\left(-C\right)=-tanC\)

\(\Leftrightarrow\frac{tanA+tanB}{1-tanA.tanB}=-tanC\Leftrightarrow tanA+tanB+tanC=tanA.tanB.tanC\left(đpcm\right)\)

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

Câu a)

Ta sử dụng 2 công thức:

∙tan(180−α)=−tanα∙tan⁡(180−α)=−tan⁡α

∙tan(α+β)=tanα+tanβ1−tanα.tanβ∙tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α.tan⁡β

Áp dụng vào bài toán:

VT=tanA+tanB+tanC=tanA+tanB+tan(180−A−B)VT=tan⁡A+tan⁡B+tan⁡C=tan⁡A+tan⁡B+tan⁡(180−A−B)

=tanA+tanB−tan(A+B)=tanA+tanB−tanA+tanB1−tanA.tanB=tan⁡A+tan⁡B−tan⁡(A+B)=tan⁡A+tan⁡B−tan⁡A+tan⁡B1−tan⁡A.tan⁡B

=(tanA+tanB)(1+11−tanA.tanB)=(tanA+tanB).−tanA.tanB1−tanA.tanB=(tan⁡A+tan⁡B)(1+11−tan⁡A.tan⁡B)=(tan⁡A+tan⁡B).−tan⁡A.tan⁡B1−tan⁡A.tan⁡B

=−tanA.tanB.tanA+tanB1−tanA.tanB=−tanA.tanB.tan(A+B)=−tan⁡A.tan⁡B.tan⁡A+tan⁡B1−tan⁡A.tan⁡B=−tan⁡A.tan⁡B.tan⁡(A+B)

=tanA.tanB.tan(180−A−B)=tan⁡A.tan⁡B.tan⁡(180−A−B)

=tanA.tanB.tanC=VP=tan⁡A.tan⁡B.tan⁡C=VP

Do đó ta có đpcm

Tam giác ABCABC có ba góc nhọn nên tanA,tanB,tanC>0tan⁡A,tan⁡B,tan⁡C>0

Áp dụng BĐT Cauchy ta có:

P=tanA+tanB+tanC≥33√tanA.tanB.tanCP=tan⁡A+tan⁡B+tan⁡C≥3tan⁡A.tan⁡B.tan⁡C3

⇔P=tanA+tanB+tanC≥33√tanA+tanB+tanC⇔P=tan⁡A+tan⁡B+tan⁡C≥3tan⁡A+tan⁡B+tan⁡C3

⇒P≥33√P⇒P≥3P3

⇒P3≥27P⇔P(P2−27)≥0⇒P3≥27P⇔P(P2−27)≥0

⇒P2−27≥0⇒P≥3√3⇒P2−27≥0⇒P≥33

Vậy Pmin=3√3Pmin=33. Dấu bằng xảy ra khi ∠A=∠B=∠C=600

Vì A, B, C là ba góc của tam giác nên ta có : A + B + C = π.

⇒ C = π - (A + B); A + B = π - C

a) Ta có: tan A + tan B + tan C = (tan A + tan B) + tan C

= tan (A + B). (1 – tan A.tan B) + tan C

= tan (π – C).(1 – tan A. tan B) + tan C

= -tan C.(1 – tan A. tan B) + tan C

= -tan C + tan A. tan B. tan C + tan C

= tan A. tan B. tan C

b) sin 2A + sin 2B + sin 2C

= 2. sin (A + B). cos (A – B) + 2.sin C. cos C

= 2. sin (π – C). cos (A – B) + 2.sin C. cos (π – (A + B))

= 2.sin C. cos (A – B) - 2.sin C. cos (A + B)

= 2.sin C.[cos (A – B) - cos (A + B)]

= 2.sin C.[-2sinA. sin(- B)]

= 2.sin C. 2.sin A. sin B ( vì sin(- B)= - sinB )

= 4. sin A. sin B. sin C