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Example 22 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Feb. 13, 2020 by Teachoo
Transcript
Example 22 Solve tan 2x = โ cot (x" + " ๐/3) tan 2x = โcot (๐ฅ" + " ๐/3) We need to make both in terms of tan Rough tan (90ยฐ + ฮธ) = โcot ฮธ โcot ฮธ = tan (90ยฐ + ฮธ) โcot ฮธ = tan (๐/2 " + ฮธ" ) Replacing ฮธ by x + ๐/3 โcot ("x + " ๐/3) = tan (๐/2 "+ x +" ๐/3) tan 2x = tan (๐/2+x" + " ๐/3) tan 2x = tan (๐/2 " + " ๐/3 " + x" ) tan 2x = tan ((3๐ + 2๐)/(2 ร 3) " + x" ) tan 2x = tan (5๐/6 " + x" ) General solution Let tan x = tan y tan 2x = tan 2y From (1) and (2) tan 2y = tan (5๐/6 " + x" ) 2y = 5๐/6 + x General solution is 2x = nฯ + 2y where n โ Z Put 2y = ("x + " 5๐/6) 2x = nฯ + ("x + " 5๐/6) 2x โ x = nฯ + 5๐/6 x = nฯ + ๐๐ /๐ where n โ Z
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Example 5 Important
Example 6 Important
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Example 20 Not in Syllabus - CBSE Exams 2021
Example 21 Not in Syllabus - CBSE Exams 2021
Example 22 Important Not in Syllabus - CBSE Exams 2021 You are here
Example 23 Not in Syllabus - CBSE Exams 2021
Example 24 Important Not in Syllabus - CBSE Exams 2021
Example 25 Important
Example 26 Important
Example 27 Important
Example 28 Important
Example 29 Important
Chapter 3 Class 11 Trigonometric Functions
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.