If A, B, C be the angles of a triangle, then which of the following hold good?

c o t A . c o t B . c o t C \leq \frac{1}{3 \sqrt{3}}

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c o t \frac{A}{2} c o t \frac{B}{2} c o t \frac{C}{2} \geq 3 \sqrt{(3)}

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c o s A + c o s B + c o s C < \frac{3}{2}

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s i n (\frac{A}{2}) s i n (\frac{B}{2}) s i n (\frac{C}{2}) \leq \frac{1}{8}

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Solution

The correct options are
A $c o t A . c o t B . c o t C \leq \frac{1}{3 \sqrt{3}}$
B $c o t \frac{A}{2} c o t \frac{B}{2} c o t \frac{C}{2} \geq 3 \sqrt{(3)}$
C $c o s A + c o s B + c o s C < \frac{3}{2}$
D $s i n (\frac{A}{2}) s i n (\frac{B}{2}) s i n (\frac{C}{2}) \leq \frac{1}{8}$
All the four hold good
(a) We know that when $A + B + C = π$ then
$t a n (A + B + C) = t a n π = 0$
or $\frac{S_{1} - S_{3}}{1 - S_{2}} = 0 ∴ S_{1} = S_{3}$
or $t a n A + t a n B + t a n C = t a n A t a n B t a n C \dots \dots (1)$
Now $t a n A + t a n B + t a n C \geq 3 (t a n A t a n B t a n C)^{\frac{1}{3}}$
or $(t a n A t a n B t a n C)^{3} \geq 27 t a n A t a n B t a n C$
or $t a n A t a n B t a n C \geq 3 \sqrt{3}$
or $c o t A c o t B c o t C \leq \frac{1}{3 \sqrt{3}} \dots \dots (2)$

(b) Again $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{π}{2}$
$∴ t a n (\frac{A}{2} + \frac{B}{2} + \frac{C}{2}) = \infty$
$= \frac{S_{1} - S_{3}}{1 - S_{2}} = \infty$
$∴ 1 - S_{2} = 0$
or $\sum t a n (\frac{A}{2}) t a n (\frac{B}{2}) = 1$
or on dividing by $t a n (\frac{A}{2}) t a n (\frac{B}{2}) t a n (\frac{C}{2})$ ,
we get $c o t (\frac{A}{2}) + c o t (\frac{B}{2}) + c o t (\frac{C}{2})$
$= c o t (\frac{A}{2}) + c o t (\frac{B}{2}) + c o t (\frac{C}{2}) \dots \dots (1)$
Now proceed as in part (a) for result (1)

(c) We have $c o s A + c o s B + c o s C$
$= 2 c o s \frac{A + B}{2} c o s \frac{A - B}{2} + c o s C$
$= 2 s i n \frac{C}{2} c o s \frac{A - B}{2} + c o s C \leq 2 s i n \frac{C}{2} + c o s C$
$[∵ 0 \leq c o s \frac{A + B}{2} \leq 1]$
$= 2 s i n \frac{C}{2} + 1 - 2 s i n^{2} \frac{C}{2}$
$= - 2 [{(s i n \frac{C}{2} - \frac{1}{2})}^{2} - \frac{1}{4}] + 1$
$= \frac{3}{2} - 2 {(s i n \frac{C}{2} - \frac{1}{2})}^{2} \leq \frac{3}{2}$

(d) We know from trigonometry that
$c o s A + c o s B + c o s C = 1 + 4 s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2} \leq \frac{3}{2}$ by (c)
$∴ 4 s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2} \leq \frac{3}{2} - 1 = \frac{1}{2}$
$∴ s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2} \leq \frac{1}{8}$

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