In a triangle ABC,tanA2tanB2tanC2^2≤

Correct answer: (A) 1/27. We know that for ∆ABC tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1 ....i For triangle, tanA/2, tanB/2, tanC/2>0 Now, AM≥GM ⇒(tanA2tanB2+tanB2tanC2+tanC2tanA2)/3≥tanA2tanB2×tanB2tanC2×tanC2tanA2^13 ⇒tan2A2tan2B2tan2C2^13≤1/3 ⇒tanA2tanB2tanC2^2≤1/27

Q61 / 160 · AP EAMCET 2022 (04 Jul Shift 2) · Mathematics

In a triangle $A B C, {(\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2})}^{2} \leq$

Correct answer
A
$\frac{1}{27}$
B
$\frac{1}{18}$
C
$\frac{1}{9}$
D
$\frac{1}{3}$

Solution

We know that for $∆ A B C$
$\tan (\frac{A}{2}) \tan (\frac{B}{2}) + \tan (\frac{B}{2}) \tan (\frac{C}{2}) + \tan (\frac{C}{2}) \tan (\frac{A}{2}) = 1 . . . . (i)$
For triangle, $\tan (\frac{A}{2}), \tan (\frac{B}{2}), \tan (\frac{C}{2}) > 0$
Now,
$AM \geq GM$
$\Rightarrow \frac{\tan (\frac{A}{2}) \tan (\frac{B}{2}) + \tan (\frac{B}{2}) \tan (\frac{C}{2}) + \tan (\frac{C}{2}) \tan (\frac{A}{2})}{3} \geq {[\tan (\frac{A}{2}) \tan (\frac{B}{2}) \times \tan (\frac{B}{2}) \tan (\frac{C}{2}) \times \tan (\frac{C}{2}) \tan (\frac{A}{2})]}^{\frac{1}{3}}$
$\Rightarrow {[\tan^{2} (\frac{A}{2}) \tan^{2} (\frac{B}{2}) \tan^{2} (\frac{C}{2})]}^{\frac{1}{3}} \leq \frac{1}{3}$
$\Rightarrow {[\tan (\frac{A}{2}) \tan (\frac{B}{2}) \tan (\frac{C}{2})]}^{2} \leq \frac{1}{27}$

Hence, the correct answer is option: 'A'.

In a triangle ABC,tanA2tanB2tanC22≤

Solution

Relevant Questions

In a triangle $A B C, {(\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2})}^{2} \leq$