If a[p]2[/p] , b[p]2[/p] , c[p]2[/p] in a ΔABC are in A.P. Prove that cot A, cot B and cot C are also in A.P.

cot A, cot B and cot C are in A.P. if :
cot A – cot B = cot B – cot C

→ cos Asin A-cos BsinB = cos BsinB - cos Csin C

→sin(B-A)sinA sinB = sin(C-B)sinC sin B

→sin (B – A) sin C = sin (C – B) sin A

→ sin (B – A) sin (B + A) = sin (C – B) sin (C + B)

→sin²B – sin²A = sin²C – sin²B

→ b²4R² - a²4R² = c²4R² - b²4R² (using sine rule)

→ b² -a² = c² - b²

→ 2 b² =a² + c²

→ a²,b²,c² are in AP

→cot A, cot B and cot C are also in A.P.

Proved

thnx