ta có: A\2+B\2 = π\2 - C\2 

⇒ tan(A\2+B\2) = tan(π\2 -C\2) 

⇒ (tanA\2 +tanB\2)\[1 - tanA\2.tanB\2] = cotgC\2 

⇒ (tanA\2 +tanB\2).tanC\2 = [1 - tanA\2.tanB\2] 

⇒ tanA\2.tanB\2 + tanB\2.tanC\2 + tanC\2.tanA\2 = 1 

............đpcm............

Tự chứng minh từng cái này rồi suy ra cái đó nhé b.

Ta có: \(sin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}-sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=sin^2\frac{A}{2}\)

Tương tự ta suy ra: 

\(sin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}+cos\frac{A}{2}sin\frac{B}{2}cos\frac{C}{2}+cos\frac{A}{2}cos\frac{B}{2}sin\frac{C}{2}=sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+3sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\left(1\right)\)

Tiếp theo chứng minh:

\(2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=\frac{cosA+cosB+cosC-1}{2}\left(2\right)\)

\(sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}=\frac{3}{2}-\frac{cosA+cosB+cosC}{2}\left(3\right)\)

\(tan\frac{A}{2}tan\frac{B}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}=1\left(4\right)\)

Từ (1), (2), (3), (4) suy được điều phải chứng minh

ko hiểu ( vì em mới học lớp 6)

trinh le na

cho bạn 4 năm nữa cũng chưa hiểu đâu

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

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

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

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

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

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

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

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

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

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

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

1.

\(sinA+sinB-sinC=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-sin\left(A+B\right)\)

\(=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-2sin\dfrac{A+B}{2}.cos\dfrac{A+B}{2}\)

\(=2sin\dfrac{A+B}{2}.\left(cos\dfrac{A-B}{2}-cos\dfrac{A+B}{2}\right)\)

\(=2sin\dfrac{A+B}{2}.2sin\dfrac{A}{2}.sin\dfrac{B}{2}\)

\(=4sin\dfrac{A}{2}.sin\dfrac{B}{2}.cos\dfrac{C}{2}\)

Sao t lại đc như này v, ai check hộ phát

a/ \(\frac{A}{2}+\left(\frac{B}{2}+\frac{C}{2}\right)=90^0\)

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

b/ \(\frac{tan^2A-tan^2B}{1-tan^2A.tan^2B}=\frac{\left(tanA-tanB\right)}{\left(1+tanA.tanB\right)}.\frac{\left(tanA+tanB\right)}{\left(1-tanA.tanB\right)}=tan\left(A-B\right).tan\left(A+B\right)\)

\(=tan\left(A-B\right).tan\left(180^0-C\right)=-tan\left(A-B\right).tanC\)

c/

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

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

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

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