{"id":9384,"date":"2022-01-11T01:39:12","date_gmt":"2022-01-10T20:09:12","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9384"},"modified":"2022-01-11T01:54:06","modified_gmt":"2022-01-10T20:24:06","slug":"prove-that-tan-a-tan-60-a-tan-60-a-3-tan-3a","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-tan-a-tan-60-a-tan-60-a-3-tan-3a\/","title":{"rendered":"Prove that tan A + tan (60 + A) –\u00a0 tan (60 – A) = 3 tan 3A."},"content":{"rendered":"

Solution :<\/h2>\n

We have,<\/p>\n

L.H.S = tan A + tan (60 + A) –  tan (60 – A)<\/p>\n

\\(\\implies\\) L.H.S = tan A + \\(\\sqrt{3} + tan A\\over 1 – \\sqrt{3} tan A\\) – \\(\\sqrt{3} – tan A\\over 1 + \\sqrt{3} tan A\\)<\/p>\n

[ By using this formula, tan (A + B) = \\(tan A + tan B\\over 1 – tan A tan B\\)  above ]<\/p>\n

\\(\\implies\\) L.H.S = tan A + \\(8 tan A\\over 1 – 3 tan^2 A\\)<\/p>\n

\\(\\implies\\) L.H.S = \\(9 tan A – 3 tan^3 A\\over 1 – 3 tan^2 A\\)<\/p>\n

L.H.S = 3(\\(3 tan A – tan^3 A\\over 1 – 3 tan^2 A\\))<\/p>\n

Since \\(3 tan A – tan^3 A\\over 1 – 3 tan^2 A\\) = tan 3A<\/p>\n

\\(\\implies\\) L.H.S = 3 tan 3A = R.H.S<\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : We have, L.H.S = tan A + tan (60 + A) –  tan (60 – A) \\(\\implies\\) L.H.S = tan A + \\(\\sqrt{3} + tan A\\over 1 – \\sqrt{3} tan A\\) – \\(\\sqrt{3} – tan A\\over 1 + \\sqrt{3} tan A\\) [ By using this formula, tan (A + B) = \\(tan A …<\/p>\n

Prove that tan A + tan (60 + A) –\u00a0 tan (60 – A) = 3 tan 3A.<\/span> Read More »<\/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":"\nProve that tan A + tan (60 + A) -\u00a0 tan (60 - A) = 3 tan 3A.<\/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-tan-a-tan-60-a-tan-60-a-3-tan-3a\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prove that tan A + tan (60 + A) -\u00a0 tan (60 - A) = 3 tan 3A.\" \/>\n<meta property=\"og:description\" content=\"Solution : We have, L.H.S = tan A + tan (60 + A) –  tan (60 – A) (implies) L.H.S = tan A + (sqrt{3} + tan Aover 1 – sqrt{3} tan A) – (sqrt{3} – tan Aover 1 + sqrt{3} tan A) [ By using this formula, tan (A + B) = (tan A … Prove that tan A + tan (60 + A) –\u00a0 tan (60 – A) = 3 tan 3A. 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