If A, B, C, are angles of triangle A B C and tan (A/2), tan (B/2), tan (C/2) are in H.P., then the minimum value of cot (B/2) is

Question

If A, B, C, are angles of triangle A B C and tan (A/2), tan (B/2), tan (C/2) are in H.P., then the minimum value of cot (B/2) is (A) -√3 (B) √3 (C) -2 √3 (D) 2 √3

Tardigrade Admin · Accepted Answer

cot ( A /2) ⋅ cot ( B /2) ⋅ cot ( C /2)= cot ( A /2)+ cot ( B /2)+ cot ( C /2) Since, cot (A/2), cot (B/2), cot (C/2) are in A.P. ⇒ cot (A/2) ⋅ cot (B/2) ⋅ cot (C/2)=3 cot (B/2) cot ( A /2) ⋅ cot ( C /2)=3 ApplyA.M. ≥ G.M. ( cot ( A /2)+ cot ( C /2)/2) ≥ √ cot ( A /2) cot ( C /2) Hence, cot (B/2) ≥ √3.

Q. If $A, B, C$ , are angles of $△ A BC$ and $tan \frac{A}{2}, tan \frac{B}{2}, tan \frac{C}{2}$ are in H.P., then the minimum value of $cot \frac{B}{2}$ is

$- 3$

$3$

$- 23$

$23$

Q. If A,B,C, are angles of △ABC and tan2A​,tan2B​,tan2C​ are in H.P., then the minimum value of cot2B​ is

−3​

3​

−23​

23​

cot2A​⋅cot2B​⋅cot2C​=cot2A​+cot2B​+cot2C​ Since, cot2A​,cot2B​,cot2C​ are in A.P. ⇒cot2A​⋅cot2B​⋅cot2C​=3cot2B​ cot2A​⋅cot2C​=3 ApplyA.M. ≥ G.M. 2cot2A​+cot2C​​≥cot2A​cot2C​​ Hence, cot2B​≥3​.

Q. If $A, B, C$ , are angles of $△ A BC$ and $tan \frac{A}{2}, tan \frac{B}{2}, tan \frac{C}{2}$ are in H.P., then the minimum value of $cot \frac{B}{2}$ is

$- 3$

$3$

$- 23$

$23$