Question Number 113822 by deepraj123 last updated on 15/Sep/20
$$\mathrm{If}\:\:\mathrm{in}\:\mathrm{a}\:\bigtriangleup{ABC},\:\mathrm{3}{a}={b}+{c},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:\mathrm{is} \\ $$
Answered by $@y@m last updated on 15/Sep/20
$$\mathrm{cot}\:^{\mathrm{2}} \frac{{B}}{\mathrm{2}}=\frac{{s}\left({s}−{b}\right)}{\left({s}−{a}\right)\left({s}−{c}\right)}\:\: {Formula} \\ $$$${Given}, \\ $$$$\mathrm{3}{a}={b}+{c} \\ $$$$\mathrm{4}{a}={a}+{b}+{c}=\mathrm{2}{s} \\ $$$${s}=\mathrm{2}{a} \\ $$$$\therefore\mathrm{cot}\:^{\mathrm{2}} \frac{{B}}{\mathrm{2}}=\frac{\mathrm{2}{a}\left(\mathrm{2}{a}−{b}\right)}{{a}\left(\mathrm{2}{a}−{c}\right)}=\frac{\mathrm{2}\left(\mathrm{2}{a}−{b}\right)}{\left(\mathrm{2}{a}−{c}\right)} \\ $$