यदि A + B + C = pi , तो {tan ^2}frac{A}{2} + | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $xy + yz + zx = 1$, 

यहाँ  $x = 	an \frac{A}{2}$ इत्यादि. 

अब ${(x - y)^2} + {(y - z)^2} + {(z - x)^2} \ge 0$

या $2\sum {x^2} - 2\sum xy \ge 0 \Rightarrow \sum {x^2} \ge 1$.   $\{ \because \sum xy = 1\} $

यदि $A + B + C = \pi ,$ तो ${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + $${\tan ^2}\frac{C}{2}$ हमेशा है

Similar Questions

यदि $\tan x + \tan \left( {\frac{\pi }{3} + x} \right) + \tan \left( {\frac{{2\pi }}{3} + x} \right) = 3,$ हो, तब

$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $

$\frac{{\cos A}}{{1 - \sin A}} = $

निम्नलिखित को सिद्ध कीजिए

$\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$

यदि $\tan \,(A + B) = p,\,\,\tan \,(A - B) = q,$ तो $\tan \,2A$ का मान $p$ तथा $q$ के पदों में है