If A + B + C = pi , then {tan ^2}frac{A}{2} + | vedclass

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Question

(b) $	an \left( {\frac{A}{2} + \frac{B}{2} + \frac{C}{2}} ight) $

$= \frac{{{S_1} - {S_3}}}{{1 - {S_2}}} = 	an \frac{\pi }{2} = \infty $

Vedclass · Accepted Answer

$ \ge 1$

Vedclass · Answer

(b) $	an \left( {\frac{A}{2} + \frac{B}{2} + \frac{C}{2}} ight) $

$= \frac{{{S_1} - {S_3}}}{{1 - {S_2}}} = 	an \frac{\pi }{2} = \infty $ 

$	herefore {S_2} = 1$ or $xy + yz + zx = 1$, 

where $x = 	an \frac{A}{2}$etc. 

Now ${(x - y)^2} + {(y - z)^2} + {(z - x)^2} \ge 0$

or $2\sum {x^2} - 2\sum xy \ge 0 \Rightarrow \sum {x^2} \ge 1$.   $\{ \because \sum xy = 1\} $

If $A + B + C = \pi ,$ then ${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + $${\tan ^2}\frac{C}{2}$ is always

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