STD 11 - 3. trignometrical ratios,functions and identities — JEE कक्षा 11 साइन्स — Question

$\therefore \,\frac{A}{2} + \frac{B}{2} = {90^o} - \frac{C}{2}$

$\therefore \cot \left( {\frac{A}{2} + \frac{B}{2}} \right) = \cot \left( {{{90}^o} - \frac{C}{2}} \right)$

या $\frac{{\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\, - 1}}{{\cot \frac{B}{2}\, + \,\cot \frac{A}{2}}} = \tan \frac{C}{2} = \frac{1}{{\cot \frac{C}{2}}}$

या $\left( {\cot \frac{A}{2}\cot \frac{B}{2} - 1} \right)\cot \frac{C}{2} = \cot \frac{B}{2} + \cot \frac{A}{2}$

$\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\,.\,\cot \frac{C}{2} = \cot \frac{C}{2} + \cot \frac{B}{2}$ $ + \cot \frac{A}{2}.$