{"id":133945,"date":"2023-03-12T16:49:08","date_gmt":"2023-03-12T11:19:08","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=133945"},"modified":"2023-03-12T16:49:08","modified_gmt":"2023-03-12T11:19:08","slug":"find-the-value-of-tan-1-2cos2sin-1-1-2tan-1-1","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/quiz\/find-the-value-of-tan-1-2cos2sin-1-1-2tan-1-1","title":{"rendered":"Find the value of tan-1 [2cos(2sin-1 1\/2)]+tan-1 1."},"content":{"rendered":"

Question: Find the value of tan-1 [2cos(2sin-1 1\/2)]+tan-1<\/h2>\n

The correct answer is –<\/h3>\n
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We can use the identity:<\/h4>\n

tan(A + B) = (tan A + tan B) \/ (1 – tan A tan B)<\/h4>\n

Let A = tan^-1[2cos(sin^-1(1\/2))] and B = tan^-1(1\/3).<\/h4>\n

Then we have:<\/h4>\n

tan A = 2cos(sin^-1(1\/2)) = 2(\u221a3 \/ 2) = \u221a3 sin A = \u221a3 \/ (sec A) = \u221a3 \/ 2<\/h4>\n

tan B = 1\/3 sin B = 1\/\u221a(1+(1\/3)^2) = 3\/\u221a10 cos B = \u221a(1 – sin^2 B) = \u221a(1 – 9\/10) = 1\/\u221a10<\/h4>\n

Using the identity above, we have:<\/h4>\n

tan(A + B) = (tan A + tan B) \/ (1 – tan A tan B) = (\u221a3 + 1\/3) \/ (1 – \u221a3\/3) = [(3\u221a3 + 1) \/ 9] \/ [(3 – \u221a3) \/ 3] = (3\u221a3 + 1) \/ (3 – \u221a3)<\/h4>\n

Now, we can simplify the expression for A + B using the formula:<\/h4>\n

tan(A + B) = (tan A + tan B) \/ (1 – tan A tan B)<\/h4>\n

Thus, we have:<\/h4>\n

tan(A + B) = (3\u221a3 + 1) \/ (3 – \u221a3)<\/h4>\n

Therefore, the value of the expression tan^-1 [2cos(sin^-1(1\/2))] + tan^-1(1\/3) is equal to the inverse tangent of (3\u221a3 + 1) \/ (3 – \u221a3), which can be simplified as follows:<\/h4>\n

tan^-1[(3\u221a3 + 1) \/ (3 – \u221a3)] = tan^-1[(3\u221a3 + 1) \/ (3 – \u221a3) * (3 + \u221a3) \/ (3 + \u221a3)] = tan^-1[(9 + 4\u221a3) \/ 6] = tan^-1[(3 + 2\u221a3) \/ 2]<\/h4>\n<\/div>\n<\/div>\n<\/div>\n
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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question: Find the value of tan-1 [2cos(2sin-1 1\/2)]+tan-1 The correct answer is – We can use the identity: tan(A + B) = (tan A + tan B) \/ (1 – tan A tan B) Let A = tan^-1[2cos(sin^-1(1\/2))] and B = tan^-1(1\/3). Then we have: tan A = 2cos(sin^-1(1\/2)) = 2(\u221a3 \/ 2) = \u221a3 […]<\/p>\n","protected":false},"author":21842,"featured_media":132233,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12119],"tags":[],"class_list":{"0":"post-133945","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-quiz"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/133945","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21842"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=133945"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/133945\/revisions"}],"predecessor-version":[{"id":133946,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/133945\/revisions\/133946"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/132233"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=133945"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=133945"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=133945"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}