{"id":127988,"date":"2019-05-16T22:12:16","date_gmt":"2019-05-17T01:12:16","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=127988"},"modified":"2023-05-31T18:47:04","modified_gmt":"2023-05-31T21:47:04","slug":"probleminha-tangente-da-soma","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/probleminha-tangente-da-soma\/","title":{"rendered":".Problem\u00e3o: Tangente da soma"},"content":{"rendered":"<p style=\"margin:30px\">\n<div class=\"g1\" style=\"line-height:25px;\">\n<p class=\"tit\" style=\"line-height:20px; margin-bottom: 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\"\/>\n<p>(<strong>Moscow Mathematical Olympiads<\/strong>, 1993-1999 &#8211; Adaptado) Sejam  [tex]\\tan  x+\\tan y=p[\/tex] e [tex]\\cot x+\\cot  y=q[\/tex], com [tex]\\tan  x, \\tan y\\neq 0[\/tex], [tex]\\tan x\\neq \\tan y[\/tex] e [tex]p\\neq q[\/tex].<br \/>\n<strong>Calcule [tex]\\tan (x+y)[\/tex].<\/strong><\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:33px\">\n<div class=\"cb36 g1n\" style=\"margin:30px;line-height:30px; width:80%; margin-left:auto; margin-right:auto; \"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" style=\"border:none\"  alt=\"explicador_p\" width=\"99\" height=\"120\" class=\"alignleft size-full wp-image-1001\" \/> <\/p>\n<p class=\"c3 ti3\" style=\"line-height:20px; margin-bottom: 10px;\"> Lembretes<\/p>\n<p>[tex]\\textcolor{#800000}{\\textbf{(i)}}[\/tex] [tex]\\boxed{\\cot a=\\dfrac{1}{\\tan a}}[\/tex], com [tex]a\\neq k\\dfrac{\\pi}{2}[\/tex], [tex]k\\in \\mathbb{Z}[\/tex].<br \/>\n[tex]\\textcolor{#800000}{\\textbf{(ii)}}[\/tex] Dados dois \u00e2ngulos [tex]a[\/tex] e [tex]b[\/tex], com [tex]a,b, a+b\\neq \\dfrac{\\pi}{2}+k\\pi[\/tex], [tex]k\\in \\mathbb{Z}[\/tex], tem-se:<br \/>\n[tex]\\qquad \\boxed{\\tan(a+b)=\\dfrac{\\tan a+\\tan b}{1-\\tan a\\cdot \\tan b}} \\, .[\/tex]<\/p>\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1\" style=\"line-height:33px;\">\n<p class=\"tit\" style=\"line-height:30px; margin-bottom: 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<p>Pelos dados do problema temos que [tex]\\cot x+\\cot y=q[\/tex]; ent\u00e3o, do [tex]\\textbf{Lembrete (i)} [\/tex], tem-se<br \/>\n[tex] \\qquad \\dfrac{1}{\\tan x}+\\dfrac{1}{\\tan y}=q \\\\<br \/>\n\\qquad \\dfrac{\\tan x+\\tan y}{\\tan x\\cdot \\tan y}=q\\\\<br \/>\n\\qquad \\tan x\\cdot \\tan y=\\dfrac{\\tan x+\\tan y}{q}.\\;\\;\\;\\;\\;\\;\\;\\; \\textcolor{#800000}{(1)}[\/tex]<br \/>\nComo [tex]\\tan x+\\tan y=p[\/tex], substituindo em [tex]\\textcolor{#800000}{(1)}[\/tex] obtemos<br \/>\n[tex] \\qquad  \\tan x\\cdot \\tan y=\\dfrac{p}{q}.\\;\\;\\;\\;\\;\\; \\textcolor{#800000}{(2)}[\/tex]<br \/>\nDo [tex]\\textbf{Lembrete (ii)}[\/tex] e da equa\u00e7\u00e3o [tex]\\textcolor{#800000}{(2)}[\/tex] temos<br \/>\n[tex] \\qquad \\tan(x+y)=\\dfrac{\\tan x+\\tan y}{1-\\tan x\\cdot \\tan y}\\\\<br \/>\n\\qquad \\tan(x+y)=\\dfrac{p}{1-\\dfrac{p}{q}}\\\\<br \/>\n\\qquad \\tan(x+y)=\\dfrac{p}{\\dfrac{q-p}{q}}\\\\<br \/>\n\\qquad  \\, \\fcolorbox{black}{#eee0e5}{$\\tan(x+y)=\\dfrac{pq}{q-p}$} \\, .[\/tex]<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">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.) (Moscow Mathematical Olympiads, 1993-1999 &#8211; Adaptado) Sejam [tex]\\tan x+\\tan y=p[\/tex] e [tex]\\cot x+\\cot y=q[\/tex], com [tex]\\tan x, \\tan y\\neq 0[\/tex], [tex]\\tan x\\neq \\tan y[\/tex] e [tex]p\\neq q[\/tex]. Calcule [tex]\\tan (x+y)[\/tex]. Lembretes [tex]\\textcolor{#800000}{\\textbf{(i)}}[\/tex] [tex]\\boxed{\\cot a=\\dfrac{1}{\\tan a}}[\/tex], com [tex]a\\neq k\\dfrac{\\pi}{2}[\/tex], [tex]k\\in \\mathbb{Z}[\/tex]. [tex]\\textcolor{#800000}{\\textbf{(ii)}}[\/tex] Dados dois \u00e2ngulos [tex]a[\/tex] &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/probleminha-tangente-da-soma\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":721,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-127988","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/127988","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\/721"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=127988"}],"version-history":[{"count":11,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/127988\/revisions"}],"predecessor-version":[{"id":192609,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/127988\/revisions\/192609"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=127988"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}