{"id":9203,"date":"2022-01-03T02:06:50","date_gmt":"2022-01-02T20:36:50","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9203"},"modified":"2022-01-03T02:13:06","modified_gmt":"2022-01-02T20:43:06","slug":"prove-that-1-tan2-theta-sec2-theta","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-1-tan2-theta-sec2-theta\/","title":{"rendered":"Prove that 1 + \\(tan^2 \\theta\\) = \\(sec^2 \\theta\\)."},"content":{"rendered":"<h2>Solution :<\/h2>\n<p>In right angled triangle ABC,<img decoding=\"async\" loading=\"lazy\" class=\"size-medium wp-image-9201 alignright\" src=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/01\/triangle-e1641154796978-300x236.png?resize=300%2C236&#038;ssl=1\" alt=\"\" width=\"300\" height=\"236\" srcset=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/01\/triangle-e1641154796978.png?resize=300%2C236&amp;ssl=1 300w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/01\/triangle-e1641154796978.png?resize=768%2C604&amp;ssl=1 768w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/01\/triangle-e1641154796978.png?w=834&amp;ssl=1 834w\" sizes=\"(max-width: 300px) 100vw, 300px\" data-recalc-dims=\"1\" \/><\/p>\n<p>\\(sec \\theta\\) = \\(AC\\over AB\\)&nbsp; \\(\\implies\\)&nbsp; &nbsp;\\(sec^2 \\theta\\) = \\(AC^2\\over AB^2\\)<\/p>\n<p>\\(tan \\theta\\) = \\(BC\\over AB\\)&nbsp; \\(\\implies\\)&nbsp; &nbsp;\\(tan^2 \\theta\\) = \\(BC^2\\over AB^2\\)<\/p>\n<p>\\(\\implies\\) 1 + \\(tan^2 \\theta\\) = 1 + \\(BC^2\\over AB^2\\)&nbsp; = \\(AB^2 + BC^2\\over AB^2\\) = \\(AC^2\\over AB^2\\)<\/p>\n<p>[ By Pythagoras theorem,&nbsp; \\(AC^2\\) = \\(BC^2 + AB^2\\) ]<\/p>\n<p>\\(\\implies\\) 1 + \\(tan^2 \\theta\\) = \\(sec^2 \\theta\\)<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : In right angled triangle ABC, \\(sec \\theta\\) = \\(AC\\over AB\\)&nbsp; \\(\\implies\\)&nbsp; &nbsp;\\(sec^2 \\theta\\) = \\(AC^2\\over AB^2\\) \\(tan \\theta\\) = \\(BC\\over AB\\)&nbsp; \\(\\implies\\)&nbsp; &nbsp;\\(tan^2 \\theta\\) = \\(BC^2\\over AB^2\\) \\(\\implies\\) 1 + \\(tan^2 \\theta\\) = 1 + \\(BC^2\\over AB^2\\)&nbsp; = \\(AB^2 + BC^2\\over AB^2\\) = \\(AC^2\\over AB^2\\) [ By Pythagoras theorem,&nbsp; \\(AC^2\\) = \\(BC^2 + &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathemerize.com\/prove-that-1-tan2-theta-sec2-theta\/\"> <span class=\"screen-reader-text\">Prove that 1 + \\(tan^2 \\theta\\) = \\(sec^2 \\theta\\).<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","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>Prove that 1 + \\(tan^2 \\theta\\) = \\(sec^2 \\theta\\). - 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\/prove-that-1-tan2-theta-sec2-theta\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prove that 1 + \\(tan^2 \\theta\\) = \\(sec^2 \\theta\\). - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : In right angled triangle ABC, (sec theta) = (ACover AB)&nbsp; (implies)&nbsp; &nbsp;(sec^2 theta) = (AC^2over AB^2) (tan theta) = (BCover AB)&nbsp; (implies)&nbsp; &nbsp;(tan^2 theta) = (BC^2over AB^2) (implies) 1 + (tan^2 theta) = 1 + (BC^2over AB^2)&nbsp; = (AB^2 + BC^2over AB^2) = (AC^2over AB^2) [ By Pythagoras theorem,&nbsp; (AC^2) = (BC^2 + &hellip; Prove that 1 + (tan^2 theta) = (sec^2 theta). 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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\/prove-that-1-tan2-theta-sec2-theta\/","og_locale":"en_US","og_type":"article","og_title":"Prove that 1 + \\(tan^2 \\theta\\) = \\(sec^2 \\theta\\). - Mathemerize","og_description":"Solution : In right angled triangle ABC, (sec theta) = (ACover AB)&nbsp; (implies)&nbsp; &nbsp;(sec^2 theta) = (AC^2over AB^2) (tan theta) = (BCover AB)&nbsp; (implies)&nbsp; &nbsp;(tan^2 theta) = (BC^2over AB^2) (implies) 1 + (tan^2 theta) = 1 + (BC^2over AB^2)&nbsp; = (AB^2 + BC^2over AB^2) = (AC^2over AB^2) [ By Pythagoras theorem,&nbsp; (AC^2) = (BC^2 + &hellip; Prove that 1 + (tan^2 theta) = (sec^2 theta). 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