Từ giả thiết ta có hệ phương trình : \(\begin{cases}\tan A.\tan B=6\\\tan A.\tan C=3\end{cases}\)

Mặt khác, ta cũng có : \(-\tan B=\tan\left(A+C\right)=\frac{\tan A+\tan C}{1-\tan A.\tan C}=\frac{\tan A+\tan C}{1-3}=-\frac{1}{2}\left(\tan A+\tan C\right)\)

\(\Leftrightarrow2\tan B=\tan A+\tan C\)

\(\Leftrightarrow2\tan A\tan B=1\tan^2A+\tan A.\tan C\)

\(\Leftrightarrow2.6=2\tan^2A+3\)

\(\Leftrightarrow\tan^2A=9\)

Theo giả thiết : \(\tan A\tan B=6>0\)

                         \(\tan A\tan C=3>0\)

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

Suy ra \(\tan A=3,\tan B=2,\tan C=1\)

Điều đó chứng tỏ \(\tan A,\tan B,\tan C\) lập thành cấp số cộng có công sai d = 1

\(\dfrac{A}{2}+\dfrac{B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}\Rightarrow tan\left(\dfrac{A}{2}+\dfrac{B}{2}\right)=tan\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)\)

\(\Rightarrow\dfrac{tan\dfrac{A}{2}+tan\dfrac{B}{2}}{1-tan\dfrac{A}{2}tan\dfrac{B}{2}}=cot\dfrac{C}{2}=\dfrac{1}{tan\dfrac{C}{2}}\)

\(\Rightarrow tan\dfrac{A}{2}.tan\dfrac{C}{2}+tan\dfrac{B}{2}tan\dfrac{C}{2}=1-tan\dfrac{A}{2}tan\dfrac{B}{2}\)

\(\Rightarrow tan\dfrac{A}{2}tan\dfrac{B}{2}+tan\dfrac{B}{2}tan\dfrac{C}{2}+tan\dfrac{C}{2}tan\dfrac{A}{2}=1\)

Ta có:

\(tan\dfrac{A}{2}+tan\dfrac{B}{2}+tan\dfrac{C}{2}\ge\sqrt{3\left(tan\dfrac{A}{2}tan\dfrac{B}{2}+tan\dfrac{B}{2}tan\dfrac{C}{2}+tan\dfrac{C}{2}tan\dfrac{A}{2}\right)}=\sqrt{3}\)

Dấu "=" xảy ra khi và chỉ khi \(A=B=C\) hay tam giác ABC đều

Đáp án D

Theo giả thiết ta có : \(\cot A+\cot C=2\cot B\)

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

\(\Leftrightarrow\sin^2B=2\sin B\sin C\cos B=\left[\cos\left(A-C\right)-\cos\left(A+C\right)\right]\cos B\)

\(\Leftrightarrow\sin^2B=\cos\left(A-C\right)\cos B-\cos\left(A+C\right)\cos B=-\cos\left(A-C\right)\cos\left(A+C\right)+\cos^2B\)

\(\Leftrightarrow\sin^2B=-\frac{1}{2}\left(\cos2A+\cos2C\right)+1-\sin^2B=-\frac{1}{2}\left(1-2\sin^2A+1-2\sin^2C\right)+1-\sin^2B\)

\(\Rightarrow2\sin^2B=\sin^2A+\sin^2C\Leftrightarrow2b^2=a^2+c^2\)

Vậy chứng tỏ \(a^2,b^2,c^2\) theo thứ tự đó cũng lập thành một cấp số cộng

Chọn A.

Ta có: a + c = 2b ⇔ sinA + sinC = 2sinB

Do đó x + y = 4.

Theo đầu bài ta có : \(\cot\frac{A}{2}+\cot\frac{C}{2}=2\cot\frac{B}{2}\Leftrightarrow\frac{\sin\frac{A+C}{2}}{\sin\frac{A}{2}\sin\frac{C}{2}}=2\frac{\cos\frac{B}{2}}{\sin\frac{B}{2}}=2\frac{\sin\frac{A+C}{2}}{\cos\frac{A+C}{2}}\)

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

\(\Leftrightarrow2\sin\frac{A+C}{2}\cos\frac{A+C}{2}=\cos\frac{A-C}{2}\sin\frac{A+C}{2}\)

\(\Leftrightarrow2\sin\left(A+C\right)=\frac{1}{2}\left(\sin A+\sin C\right)\)

\(\Leftrightarrow\sin A+\sin C=2\sin B\Rightarrow a+c=2b\)

Chứng tỏ 3 cạnh của tam giác lập thành cấp số cộng

gọi a,b,c là 3 cạnh của tam giác.

Ta có :\(cot\left(\dfrac{A}{2}\right)+cot\left(\dfrac{C}{2}\right)=2cot\left(\dfrac{B}{2}\right)\) <=> \(\dfrac{cot\left(\dfrac{A}{2}\right)}{sin\left(\dfrac{A}{2}\right)}+\dfrac{cos\left(\dfrac{C}{2}\right)}{sin\left(\dfrac{C}{2}\right)}=\dfrac{2.cos\left(\dfrac{B}{2}\right)}{sin\left(\dfrac{B}{2}\right)}\)

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

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

<=> \(sin\left(\dfrac{B}{2}\right).cos\left(\dfrac{B}{2}\right)=2sin\left(\dfrac{A}{2}\right)sin\left(\dfrac{C}{2}\right)cos\left(\dfrac{B}{2}\right)\)

<=> \(\dfrac{1}{2}sinB=\left[cos\left(\dfrac{A}{2}-\dfrac{C}{2}\right)-cos\left(\dfrac{A}{2}+\dfrac{C}{2}\right)\right]cos\left(\dfrac{B}{2}\right)\)

<=>\(\dfrac{1}{2}sinB=cos\left(\dfrac{A}{2}-\dfrac{C}{2}\right).cos\left(\dfrac{B}{2}\right)-sin\left(\dfrac{B}{2}\right)cos\left(\dfrac{B}{2}\right)\)

<=> \(\dfrac{1}{2}sinB=cos\left(\dfrac{A}{2}-\dfrac{C}{2}\right)sin\left(\dfrac{A}{2}+\dfrac{C}{2}\right)-\dfrac{1}{2}sinB\)

<=> sinB = \(\dfrac{1}{2}\left(sinA+sinC\right)\) <=> \(2sinB=sinA+sinC\)

<=> \(2.\dfrac{b}{2R}=\dfrac{a}{2R}+\dfrac{c}{2R}\)

<=> a+c =2b

=> 3 cạnh của tam giác tạo thành cấp số cộng.