Estude! Com Prof. Caju

Em um triângulo, retângulo em [tex3]A[/tex3], calcular [tex3]tg(A+B-2C)[/tex3] em função dos catetos [tex3]b[/tex3] e [tex3]c[/tex3].

Olá, Aldrin.

[tex3]tg(A+B-2C)=tg(90+B-2C)=tg[(90^\circ-C)+(B-C)][/tex3]

[tex3]tg(90^\circ-C)=\frac{tg90^\circ-tgC}{1+tg90^\circ.tgC}[/tex3]

[tex3]tg(B-C)=\frac{tgB-tgC}{1+tgB.tgC}[/tex3]

[tex3]tg[(90^\circ-C)+(B-C)]=\frac{\frac{tg90^\circ-tgC}{1+tg90^\circ.tgC}+\frac{tgB-tgC}{1+tgB.tgC}}{1-[\frac{tg90^\circ-tgC}{1+tg90^\circ.tgC}.\frac{tgB-tgC}{1+tgB.tgC}}[/tex3]

[tex3]\frac{tg90^\circ-tgC}{1+tg90^\circ.tgC}=\frac{0-\frac{c}{b}}{1+0.\frac{c}{b}}=-\frac{c}{b}[/tex3]

[tex3]\frac{tgB-tgC}{1+tgB.tgC}=\frac{\frac{b}{c}-\frac{c}{b}}{1+\frac{b}{c}.\frac{c}{b}}=\frac{\frac{b^2-c^2}{bc}}{2}=\frac{b^2-c^2}{2bc}[/tex3]

[tex3]tg[(90^\circ-C)+(B-C)]=\frac{-\frac{c}{b}+\frac{b^2-c^2}{2bc}}{1-[-\frac{c}{b}.\frac{(b^2-c^2}{2bc}]}=\frac{\frac{-2c^2+(b^2-c^2)}{2bc}}{1+\frac{(b^2-c^2)}{2b^2}}=\frac{\frac{-3c^2+b^2}{2bc}}{\frac{2b^2+b^2-c^2}{2b^2}} =[/tex3]

[tex3]=\frac{-3c^2+b^2}{2bc}.\frac{2b^2}{3b^2-c^2}=\frac{b}{c}.\frac{-3c^2+b^2}{3b^2-c^2}[/tex3]

Multiplicando o numerador e denominador por [tex3](-1)[/tex3] teremos:

[tex3]\frac{b}{c}.\frac{3c^2-b^2}{c^2-3b^2}[/tex3]