Question-22929

Question Number 22929 by selestian last updated on 24/Oct/17

Answered by ajfour last updated on 24/Oct/17

(A) r(cot (A/2)+cot (B/2))=c r(cot (B/2)+cot (C/2))=a r(cot (C/2)+cot (A/2))=b 2b=a+c , so 2r(cot (C/2)+cot (A/2))=r(cot (A/2)+ +2cot (B/2)+cot (C/2)) ⇒ cot (C/2)+cot (A/2)=2cot (B/2) For a triangle Σcot (A/2)=Πcot (A/2) ⇒cot (A/2)cot (B/2)cot (C/2) −cot (B/2) =2cot (B/2) ⇒ cot (A/2)cot (C/2)=3 ⇒ cot (A/2)=3tan (A/2) .

$$\left({A}\right) \\ $$$${r}\left(\mathrm{cot}\:\frac{{A}}{\mathrm{2}}+\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\right)={c} \\ $$$${r}\left(\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right)={a} \\ $$$${r}\left(\mathrm{cot}\:\frac{{C}}{\mathrm{2}}+\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\right)={b} \\ $$$$\mathrm{2}{b}={a}+{c}\:,\:{so} \\ $$$$\mathrm{2}{r}\left(\mathrm{cot}\:\frac{{C}}{\mathrm{2}}+\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\right)={r}\left(\mathrm{cot}\:\frac{{A}}{\mathrm{2}}+\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}+\mathrm{cot}\:\frac{{A}}{\mathrm{2}}=\mathrm{2cot}\:\frac{{B}}{\mathrm{2}} \\ $$$${For}\:{a}\:{triangle}\:\Sigma\mathrm{cot}\:\frac{{A}}{\mathrm{2}}=\Pi\mathrm{cot}\:\frac{{A}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\:−\mathrm{cot}\:\frac{{B}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2cot}\:\frac{{B}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\mathrm{cot}\:\frac{{C}}{\mathrm{2}}=\mathrm{3} \\ $$$$\Rightarrow\:\:\:\:\mathrm{cot}\:\frac{\boldsymbol{{A}}}{\mathrm{2}}=\mathrm{3tan}\:\frac{{A}}{\mathrm{2}}\:. \\ $$$$\:\:\:\: \\ $$$$ \\ $$

Commented by selestian last updated on 24/Oct/17