Question 5 - Examples - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Question 5 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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo