{"id":11911,"date":"2022-09-03T21:38:26","date_gmt":"2022-09-03T16:08:26","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11911"},"modified":"2022-09-03T21:38:30","modified_gmt":"2022-09-03T16:08:30","slug":"write-the-other-trigonometric-ratios-of-a-in-terms-of-sec-a","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/write-the-other-trigonometric-ratios-of-a-in-terms-of-sec-a\/","title":{"rendered":"Write the other trigonometric ratios of A in terms of sec A."},"content":{"rendered":"<h2>Solution :<\/h2>\n<p>Consider a triangle ABC, in which \\(\\angle\\) B = 90<img decoding=\"async\" loading=\"lazy\" class=\" wp-image-11912 alignright\" src=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/09\/trigonometry8-4-2.jpeg?resize=277%2C181&#038;ssl=1\" alt=\"triangle\" width=\"277\" height=\"181\" srcset=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/09\/trigonometry8-4-2.jpeg?resize=300%2C196&amp;ssl=1 300w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/09\/trigonometry8-4-2.jpeg?resize=1024%2C670&amp;ssl=1 1024w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/09\/trigonometry8-4-2.jpeg?resize=768%2C502&amp;ssl=1 768w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/09\/trigonometry8-4-2.jpeg?resize=1536%2C1005&amp;ssl=1 1536w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/09\/trigonometry8-4-2.jpeg?w=1700&amp;ssl=1 1700w\" sizes=\"(max-width: 277px) 100vw, 277px\" data-recalc-dims=\"1\" \/><\/p>\n<p>For \\(\\angle\\) A, we have :<\/p>\n<p>Base = AB,<\/p>\n<p>Perp = BC<\/p>\n<p>and Hyp = AC<\/p>\n<p>\\(\\therefore\\)\u00a0 \u00a0 sec A = \\(Hyp\\over Base\\) = \\(AC\\over AB\\)<\/p>\n<p>So,\u00a0 \\(AC\\over AB\\)\u00a0 = sec A = \\(sec A\\over 1\\)<\/p>\n<p>Let AB = k and AC = k sec A<\/p>\n<p>So,\u00a0 BC = \\(\\sqrt{{AC}^2 &#8211; {AB}^2}\\) = \\(k\\sqrt{sec^2 A &#8211; 1}\\)<\/p>\n<p>Now, sin A = \\(BC\\over AB\\) = <strong>\\(\\sqrt{sec^2 A &#8211; 1}\\over sec A\\)<\/strong><\/p>\n<p>cos A = \\(AB\\over AC\\) = <strong>\\(1\\over sec A\\)<\/strong><\/p>\n<p>tan A = \\(BC\\over AB\\) = <strong>\\(\\sqrt{sec^2 A &#8211; 1}\\)<\/strong><\/p>\n<p>cot A = \\(1\\over tan A\\) = <strong>\\(1\\over \\sqrt{sec^2 A &#8211; 1}\\)<\/strong><\/p>\n<p>cosec A = \\(1\\over sin A\\) = <strong>\\(sec A\\over \\sqrt{sec^2 A &#8211; 1}\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : Consider a triangle ABC, in which \\(\\angle\\) B = 90 For \\(\\angle\\) A, we have : Base = AB, Perp = BC and Hyp = AC \\(\\therefore\\)\u00a0 \u00a0 sec A = \\(Hyp\\over Base\\) = \\(AC\\over AB\\) So,\u00a0 \\(AC\\over AB\\)\u00a0 = sec A = \\(sec A\\over 1\\) Let AB = k and AC = &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathemerize.com\/write-the-other-trigonometric-ratios-of-a-in-terms-of-sec-a\/\"> <span class=\"screen-reader-text\">Write the other trigonometric ratios of A in terms of sec A.<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","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":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,60],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Write the other trigonometric ratios of A in terms of sec A. - Mathemerize<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/write-the-other-trigonometric-ratios-of-a-in-terms-of-sec-a\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Write the other trigonometric ratios of A in terms of sec A. - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Consider a triangle ABC, in which (angle) B = 90 For (angle) A, we have : Base = AB, Perp = BC and Hyp = AC (therefore)\u00a0 \u00a0 sec A = (Hypover Base) = (ACover AB) So,\u00a0 (ACover AB)\u00a0 = sec A = (sec Aover 1) Let AB = k and AC = &hellip; Write the other trigonometric ratios of A in terms of sec A. 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Mathemerize","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mathemerize.com\/write-the-other-trigonometric-ratios-of-a-in-terms-of-sec-a\/","og_locale":"en_US","og_type":"article","og_title":"Write the other trigonometric ratios of A in terms of sec A. - Mathemerize","og_description":"Solution : Consider a triangle ABC, in which (angle) B = 90 For (angle) A, we have : Base = AB, Perp = BC and Hyp = AC (therefore)\u00a0 \u00a0 sec A = (Hypover Base) = (ACover AB) So,\u00a0 (ACover AB)\u00a0 = sec A = (sec Aover 1) Let AB = k and AC = &hellip; Write the other trigonometric ratios of A in terms of sec A. 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