Question

The minimum value of tan B tan C in an acute angled triangle ABC is-

• A.
• B. $\cot \frac { \mathrm { A } } { 2 }$
• C. $\operatorname { cosec } ^ { 2 } \frac { \mathrm { A } } { 2 }$
• D.

Answer 1

Answer: (D)

Answer 2

Let x = tanB tanC

tanA = $\frac { \tan B + \tan C } { x - 1 } \geq \frac { 2 \sqrt { x } } { x - 1 }$ by A.M. ≥ G.M.

⇒ x – 2cotA . – 1 ≥ 0

⇒ ( – cotA)² – cosec²A ≥ 0

⇒ $\sqrt { \mathrm { x } }$ – cotA – cosecA ≥ 0 ⇒ x ≥ $\cot ^ { 2 } \frac { \mathrm { A } } { 2 }$