{"id":230708,"date":"2025-06-10T03:54:21","date_gmt":"2025-06-10T03:54:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=230708"},"modified":"2025-06-10T03:54:23","modified_gmt":"2025-06-10T03:54:23","slug":"if-sin-t-3-5-and-the-terminal-point-ofis-in-quadrant-ii-then-cos-t-tan-t-cot-t-sec-t-csc-t","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/10\/if-sin-t-3-5-and-the-terminal-point-ofis-in-quadrant-ii-then-cos-t-tan-t-cot-t-sec-t-csc-t\/","title":{"rendered":"If \\sin t = 3\/5 and the terminal point ofis in quadrant II, then \\cos t = \\tan t = \\cot t = \\sec t = \\csc t ="},"content":{"rendered":"\n

If \\sin t = 3\/5 and the terminal point of
is in quadrant II, then \\cos t = \\tan t = \\cot t = \\sec t = \\csc t =<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Given:
sin\u2061t=35\\sin t = \\dfrac{3}{5}sint=53\u200b, and the terminal point of ttt is in Quadrant II<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Use the Pythagorean identity to find cos\u2061t\\cos tcost<\/h3>\n\n\n\n

The identity is:sin\u20612t+cos\u20612t=1\\sin^2 t + \\cos^2 t = 1sin2t+cos2t=1<\/p>\n\n\n\n

Substitute sin\u2061t=35\\sin t = \\dfrac{3}{5}sint=53\u200b:(35)2+cos\u20612t=1925+cos\u20612t=1cos\u20612t=1\u2212925=1625\\left(\\dfrac{3}{5}\\right)^2 + \\cos^2 t = 1 \\\\ \\dfrac{9}{25} + \\cos^2 t = 1 \\\\ \\cos^2 t = 1 – \\dfrac{9}{25} = \\dfrac{16}{25}(53\u200b)2+cos2t=1259\u200b+cos2t=1cos2t=1\u2212259\u200b=2516\u200b<\/p>\n\n\n\n

So,cos\u2061t=\u00b145\\cos t = \\pm \\dfrac{4}{5}cost=\u00b154\u200b<\/p>\n\n\n\n

But since the angle is in Quadrant II<\/strong>, and cosine is negative<\/strong> in Quadrant II:cos\u2061t=\u221245\\cos t = -\\dfrac{4}{5}cost=\u221254\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 2: Find the remaining trig functions<\/h3>\n\n\n\n

Now that we have:<\/p>\n\n\n\n

    \n
  • sin\u2061t=35\\sin t = \\dfrac{3}{5}sint=53\u200b<\/li>\n\n\n\n
  • cos\u2061t=\u221245\\cos t = -\\dfrac{4}{5}cost=\u221254\u200b<\/li>\n<\/ul>\n\n\n\n

    We can compute:<\/p>\n\n\n\n

      \n
    1. tan\u2061t=sin\u2061tcos\u2061t\\tan t = \\dfrac{\\sin t}{\\cos t}tant=costsint\u200b<\/strong> tan\u2061t=3\/5\u22124\/5=\u221234\\tan t = \\dfrac{3\/5}{-4\/5} = -\\dfrac{3}{4}tant=\u22124\/53\/5\u200b=\u221243\u200b<\/li>\n\n\n\n
    2. cot\u2061t=1tan\u2061t=1\u22123\/4=\u221243\\cot t = \\dfrac{1}{\\tan t} = \\dfrac{1}{-3\/4} = -\\dfrac{4}{3}cott=tant1\u200b=\u22123\/41\u200b=\u221234\u200b<\/strong><\/li>\n\n\n\n
    3. sec\u2061t=1cos\u2061t=1\u22124\/5=\u221254\\sec t = \\dfrac{1}{\\cos t} = \\dfrac{1}{-4\/5} = -\\dfrac{5}{4}sect=cost1\u200b=\u22124\/51\u200b=\u221245\u200b<\/strong><\/li>\n\n\n\n
    4. csc\u2061t=1sin\u2061t=13\/5=53\\csc t = \\dfrac{1}{\\sin t} = \\dfrac{1}{3\/5} = \\dfrac{5}{3}csct=sint1\u200b=3\/51\u200b=35\u200b<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      Final Answers:<\/h3>\n\n\n\n
        \n
      • sin\u2061t=35\\sin t = \\dfrac{3}{5}sint=53\u200b<\/li>\n\n\n\n
      • cos\u2061t=\u221245\\cos t = -\\dfrac{4}{5}cost=\u221254\u200b<\/li>\n\n\n\n
      • tan\u2061t=\u221234\\tan t = -\\dfrac{3}{4}tant=\u221243\u200b<\/li>\n\n\n\n
      • cot\u2061t=\u221243\\cot t = -\\dfrac{4}{3}cott=\u221234\u200b<\/li>\n\n\n\n
      • sec\u2061t=\u221254\\sec t = -\\dfrac{5}{4}sect=\u221245\u200b<\/li>\n\n\n\n
      • csc\u2061t=53\\csc t = \\dfrac{5}{3}csct=35\u200b<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Explanation <\/h3>\n\n\n\n

        To solve this trigonometric problem, we start with the given value of sin\u2061t=35\\sin t = \\dfrac{3}{5}sint=53\u200b, meaning the opposite side of a right triangle is 3 and the hypotenuse is 5. We need to find all other trigonometric functions based on this.<\/p>\n\n\n\n

        Using the Pythagorean identity, sin\u20612t+cos\u20612t=1\\sin^2 t + \\cos^2 t = 1sin2t+cos2t=1, we can find cos\u2061t\\cos tcost. Substituting sin\u2061t=35\\sin t = \\dfrac{3}{5}sint=53\u200b, we calculate:cos\u20612t=1\u2212(35)2=1\u2212925=1625\\cos^2 t = 1 – \\left(\\dfrac{3}{5}\\right)^2 = 1 – \\dfrac{9}{25} = \\dfrac{16}{25}cos2t=1\u2212(53\u200b)2=1\u2212259\u200b=2516\u200b<\/p>\n\n\n\n

        Taking the square root gives cos\u2061t=\u00b145\\cos t = \\pm \\dfrac{4}{5}cost=\u00b154\u200b. However, the quadrant of the angle matters. In Quadrant II<\/strong>, sine is positive but cosine is negative. So, cos\u2061t=\u221245\\cos t = -\\dfrac{4}{5}cost=\u221254\u200b.<\/p>\n\n\n\n

        Now that both sine and cosine are known, we can derive the remaining trigonometric functions:<\/p>\n\n\n\n

          \n
        • tan\u2061t=sin\u2061tcos\u2061t=3\/5\u22124\/5=\u221234\\tan t = \\dfrac{\\sin t}{\\cos t} = \\dfrac{3\/5}{-4\/5} = -\\dfrac{3}{4}tant=costsint\u200b=\u22124\/53\/5\u200b=\u221243\u200b<\/li>\n\n\n\n
        • cot\u2061t=1tan\u2061t=\u221243\\cot t = \\dfrac{1}{\\tan t} = -\\dfrac{4}{3}cott=tant1\u200b=\u221234\u200b<\/li>\n\n\n\n
        • sec\u2061t=1cos\u2061t=\u221254\\sec t = \\dfrac{1}{\\cos t} = -\\dfrac{5}{4}sect=cost1\u200b=\u221245\u200b<\/li>\n\n\n\n
        • csc\u2061t=1sin\u2061t=53\\csc t = \\dfrac{1}{\\sin t} = \\dfrac{5}{3}csct=sint1\u200b=35\u200b<\/li>\n<\/ul>\n\n\n\n

          Understanding quadrant signs is crucial: in Quadrant II, only sine and cosecant are positive. The other functions are negative due to the signs of x and y coordinates in that quadrant.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          If \\sin t = 3\/5 and the terminal point ofis in quadrant II, then \\cos t = \\tan t = \\cot t = \\sec t = \\csc t = The Correct Answer and Explanation is: Given:sin\u2061t=35\\sin t = \\dfrac{3}{5}sint=53\u200b, and the terminal point of ttt is in Quadrant II. Step 1: Use the Pythagorean identity […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-230708","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230708","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=230708"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230708\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230708"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230708"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230708"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}