{"id":45526,"date":"2015-09-11T17:38:12","date_gmt":"2015-09-11T20:38:12","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=45526"},"modified":"2023-03-29T03:25:50","modified_gmt":"2023-03-29T06:25:50","slug":"tangente-e-tangentes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/tangente-e-tangentes\/","title":{"rendered":".Problem\u00e3o: Tangente e tangentes"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nSabendo que [tex]\\,\\boxed{tg (\\alpha) + tg(\\beta)=p}\\;[\/tex] e [tex]\\;\\boxed{cotg(\\alpha) + cotg(\\beta)=q}[\/tex], determine [tex]\\,tg(\\alpha + \\beta)[\/tex].<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n Caso esteja definida [tex]tg( \\alpha + \\beta)[\/tex], ent\u00e3o:<\/p>\n<p style=\"margin:10px\">\n[tex]\\quad tg(\\alpha + \\beta)=\\dfrac{tg(\\alpha)+tg(\\beta)}{1-tg(\\alpha) \\cdot tg(\\beta)}=\\dfrac{p}{1-tg(\\alpha) \\cdot tg(\\beta)}\\,.\\qquad [\/tex] <strong><em>(1)<\/em><\/strong><\/p>\n<p>Al\u00e9m disso, temos que:<\/p>\n<p style=\"margin:10px\">\n[tex]\\quad q = cotg(\\alpha) + cotg(\\beta)= \\dfrac{1}{tg(\\alpha)}+\\dfrac{1}{tg(\\beta)}=\\dfrac{tg(\\alpha)+tg(\\beta)}{tg(\\alpha) \\cdot tg(\\beta)}=\\dfrac{p}{tg(\\alpha) \\cdot tg(\\beta)}\\,.\\qquad [\/tex] <strong><em>(2)<\/em><\/strong><\/p>\n<p>Da equa\u00e7\u00e3o <strong><em>(2)<\/em><\/strong> podemos deduzir que ou [tex]\\,p\\,[\/tex] e [tex]\\,q\\,[\/tex] s\u00e3o ambos iguais a zero ou ambos diferentes de zero.\n<\/p>\n<p><strong>[tex]1^o[\/tex] Caso:<\/strong> Se [tex]p=q=0[\/tex], ent\u00e3o de  <strong><em>(1)<\/em><\/strong> temos [tex]\\fcolorbox{black}{#eee0e5}{$tg(\\alpha+\\beta)=0$}[\/tex]. Aqui, faz-se necess\u00e1rio verificar se o denominador de  <strong><em>(1)<\/em><\/strong> n\u00e3o se anula.<br \/>\nDe fato, como [tex]p=0[\/tex], temos [tex]tg(\\alpha)=- tg(\\beta)[\/tex] e, portanto, [tex]~1-tg(\\alpha) \\cdot tg(\\beta)=1+ tg^2(\\alpha)>0.[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p><strong>[tex]2^o[\/tex] Caso:<\/strong> Se [tex]p \\neq 0[\/tex], [tex]q \\neq 0[\/tex] e [tex]p \\neq q[\/tex], ent\u00e3o <strong><em>(2)<\/em><\/strong> indica que [tex]tg (\\alpha) \\cdot tg(\\beta)=\\dfrac{p}{q}[\/tex] o que, junto com <strong><em>(1)<\/em><\/strong>, implica que [tex]\\fcolorbox{black}{#eee0e5}{$tg( \\alpha + \\beta)=\\dfrac{p\\,q}{q-p}$}[\/tex].<\/p>\n<p style=\"margin:10px\">\n<p><strong>[tex]3^o[\/tex] Caso:<\/strong> Se [tex]p \\neq 0[\/tex], [tex]q \\neq 0[\/tex] e [tex]p = q[\/tex], ent\u00e3o [tex]tg(\\alpha + \\beta)[\/tex] n\u00e3o est\u00e1 definida, pois ter\u00edamos, a partir de <strong><em>(2)<\/em><\/strong>, que [tex]tg(\\alpha) \\cdot tg(\\beta)=1[\/tex] e isso faria com que o segundo membro de <strong><em>(1)<\/em><\/strong> tivesse denominador nulo.<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:10px\">\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 2\u00ba ano do E. M.) Sabendo que [tex]\\,\\boxed{tg (\\alpha) + tg(\\beta)=p}\\;[\/tex] e [tex]\\;\\boxed{cotg(\\alpha) + cotg(\\beta)=q}[\/tex], determine [tex]\\,tg(\\alpha + \\beta)[\/tex]. Solu\u00e7\u00e3o Caso esteja definida [tex]tg( \\alpha + \\beta)[\/tex], ent\u00e3o: [tex]\\quad tg(\\alpha + \\beta)=\\dfrac{tg(\\alpha)+tg(\\beta)}{1-tg(\\alpha) \\cdot tg(\\beta)}=\\dfrac{p}{1-tg(\\alpha) \\cdot tg(\\beta)}\\,.\\qquad [\/tex] (1) Al\u00e9m disso, temos que: [tex]\\quad q = cotg(\\alpha) + cotg(\\beta)= \\dfrac{1}{tg(\\alpha)}+\\dfrac{1}{tg(\\beta)}=\\dfrac{tg(\\alpha)+tg(\\beta)}{tg(\\alpha) \\cdot &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/tangente-e-tangentes\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":166,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-45526","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/45526","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/166"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=45526"}],"version-history":[{"count":27,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/45526\/revisions"}],"predecessor-version":[{"id":185719,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/45526\/revisions\/185719"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=45526"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}