what is tan(-5\u03c0\/6)-cot(7P\u03c0\/2) without a calculator<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s evaluate the expression ( \\tan \\left( -\\frac{5\\pi}{6} \\right) – \\cot \\left( \\frac{7\\pi}{2} \\right) ) step by step.<\/p>\n\n\n\n We start by simplifying the tangent term. The tangent function has a periodicity of ( \\pi ), meaning that ( \\tan(\\theta) = \\tan(\\theta + n\\pi) ), where ( n ) is any integer.<\/p>\n\n\n\n First, we find a coterminal angle for ( -\\frac{5\\pi}{6} ) by adding ( 2\\pi ) (since ( 2\\pi ) is one full revolution):<\/p>\n\n\n\n [ Now, we know that ( \\tan \\left( -\\frac{5\\pi}{6} \\right) = \\tan \\left( \\frac{7\\pi}{6} \\right) ), since the tangent function has a period of ( \\pi ).<\/p>\n\n\n\n The angle ( \\frac{7\\pi}{6} ) is in the third quadrant (since ( \\frac{7\\pi}{6} > \\pi ), but less than ( \\frac{3\\pi}{2} )), where the tangent function is positive. The reference angle is:<\/p>\n\n\n\n [ Since ( \\tan(\\frac{\\pi}{6}) = \\frac{1}{\\sqrt{3}} ), we have:<\/p>\n\n\n\n [ Thus, ( \\tan \\left( -\\frac{5\\pi}{6} \\right) = \\frac{1}{\\sqrt{3}} ).<\/p>\n\n\n\n The cotangent function has a periodicity of ( \\pi ), meaning that ( \\cot(\\theta) = \\cot(\\theta + n\\pi) ).<\/p>\n\n\n\n Now, we simplify ( \\frac{7\\pi}{2} ) by subtracting ( 2\\pi ) (which is equivalent to subtracting ( \\frac{4\\pi}{2} )) to bring it within one full cycle:<\/p>\n\n\n\n [ Thus, ( \\cot \\left( \\frac{7\\pi}{2} \\right) = \\cot \\left( \\frac{3\\pi}{2} \\right) ).<\/p>\n\n\n\n The angle ( \\frac{3\\pi}{2} ) lies on the negative ( y )-axis. Since cotangent is the reciprocal of tangent, and ( \\tan \\left( \\frac{3\\pi}{2} \\right) = 0 ), we have:<\/p>\n\n\n\n [ This is undefined.<\/p>\n\n\n\n Since ( \\cot \\left( \\frac{7\\pi}{2} \\right) ) is undefined, the entire expression ( \\tan \\left( -\\frac{5\\pi}{6} \\right) – \\cot \\left( \\frac{7\\pi}{2} \\right) ) is undefined.<\/p>\n\n\n\n Thus, the correct answer is that the expression is undefined<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" what is tan(-5\u03c0\/6)-cot(7P\u03c0\/2) without a calculator The Correct Answer and Explanation is : Let’s evaluate the expression ( \\tan \\left( -\\frac{5\\pi}{6} \\right) – \\cot \\left( \\frac{7\\pi}{2} \\right) ) step by step. Step 1: Evaluate ( \\tan \\left( -\\frac{5\\pi}{6} \\right) ) We start by simplifying the tangent term. The tangent function has a periodicity of ( […]<\/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-188901","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188901","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=188901"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188901\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188901"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188901"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188901"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Evaluate ( \\tan \\left( -\\frac{5\\pi}{6} \\right) )<\/h3>\n\n\n\n
Finding a Coterminal Angle<\/h4>\n\n\n\n
-\\frac{5\\pi}{6} + 2\\pi = -\\frac{5\\pi}{6} + \\frac{12\\pi}{6} = \\frac{7\\pi}{6}
]<\/p>\n\n\n\nEvaluating ( \\tan \\left( \\frac{7\\pi}{6} \\right) )<\/h4>\n\n\n\n
\\frac{7\\pi}{6} – \\pi = \\frac{\\pi}{6}
]<\/p>\n\n\n\n
\\tan \\left( \\frac{7\\pi}{6} \\right) = \\frac{1}{\\sqrt{3}}
]<\/p>\n\n\n\nStep 2: Evaluate ( \\cot \\left( \\frac{7\\pi}{2} \\right) )<\/h3>\n\n\n\n
Finding a Coterminal Angle<\/h4>\n\n\n\n
\\frac{7\\pi}{2} – 2\\pi = \\frac{7\\pi}{2} – \\frac{4\\pi}{2} = \\frac{3\\pi}{2}
]<\/p>\n\n\n\nEvaluating ( \\cot \\left( \\frac{3\\pi}{2} \\right) )<\/h4>\n\n\n\n
\\cot \\left( \\frac{3\\pi}{2} \\right) = \\frac{1}{0}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n