From tan theta = sqrt 3 , how do we arrive at 4pi\/3? I did theta = inverse tan sqrt3 and did not get 4pi\/3<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve for ( \\theta ) when given that ( \\tan(\\theta) = \\sqrt{3} ), the process involves understanding both the inverse tangent function and the nature of the tangent function. Let’s break this down step by step:<\/p>\n\n\n\n The equation you’re working with is:<\/p>\n\n\n\n [ We want to find ( \\theta ). To do this, we can apply the inverse tangent function (also called ( \\tan^{-1} ) or arctan) to both sides of the equation.<\/p>\n\n\n\n [ The arctangent function ( \\tan^{-1}(\\sqrt{3}) ) gives the principal value of ( \\theta ), which is the value of ( \\theta ) in the interval ( \\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right) ).<\/p>\n\n\n\n Since ( \\tan^{-1}(\\sqrt{3}) = \\frac{\\pi}{3} ), we know that:<\/p>\n\n\n\n [ However, this only gives one solution within the principal range.<\/p>\n\n\n\n The tangent function has a period of ( \\pi ), meaning that it repeats every ( \\pi ) radians. This implies that if ( \\tan(\\theta) = \\sqrt{3} ) at ( \\theta = \\frac{\\pi}{3} ), then ( \\tan(\\theta) ) will also be ( \\sqrt{3} ) at ( \\theta = \\frac{\\pi}{3} + n\\pi ), where ( n ) is any integer.<\/p>\n\n\n\n To find all possible values of ( \\theta ) that satisfy ( \\tan(\\theta) = \\sqrt{3} ), we write the general solution as:<\/p>\n\n\n\n [ If you’re asked to find the solution in a specific interval, such as ( 0 \\leq \\theta < 2\\pi ), you would need to consider the values of ( n ) that satisfy this range.<\/p>\n\n\n\n Thus, the two solutions in the interval ( 0 \\leq \\theta < 2\\pi ) are:<\/p>\n\n\n\n [ The value ( \\theta = \\frac{4\\pi}{3} ) comes from considering the periodic nature of the tangent function. While the inverse tangent gives you ( \\frac{\\pi}{3} ), the tangent function repeats every ( \\pi ) radians, so ( \\frac{4\\pi}{3} ) is another valid solution in the interval ( 0 \\leq \\theta < 2\\pi ).<\/p>\n","protected":false},"excerpt":{"rendered":" From tan theta = sqrt 3 , how do we arrive at 4pi\/3? I did theta = inverse tan sqrt3 and did not get 4pi\/3 The Correct Answer and Explanation is : To solve for ( \\theta ) when given that ( \\tan(\\theta) = \\sqrt{3} ), the process involves understanding both the inverse tangent function […]<\/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-195647","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195647","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=195647"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195647\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195647"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195647"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195647"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Understand the given equation.<\/h3>\n\n\n\n
\\tan(\\theta) = \\sqrt{3}
]<\/p>\n\n\n\nStep 2: Apply the inverse tangent function.<\/h3>\n\n\n\n
\\theta = \\tan^{-1}(\\sqrt{3})
]<\/p>\n\n\n\n
\\theta = \\frac{\\pi}{3}
]<\/p>\n\n\n\nStep 3: Consider the periodic nature of the tangent function.<\/h3>\n\n\n\n
Step 4: Find the general solution.<\/h3>\n\n\n\n
\\theta = \\frac{\\pi}{3} + n\\pi
]<\/p>\n\n\n\nStep 5: Consider all possible solutions within a specific range.<\/h3>\n\n\n\n
\n
\\theta = \\frac{\\pi}{3}, \\frac{4\\pi}{3}
]<\/p>\n\n\n\nStep 6: Conclusion<\/h3>\n\n\n\n