Example 22 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
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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Chapter 3 Class 11 Trigonometric Functions (Term 2)
Serial order wise
- Ex 3.1
- Ex 3.2
- Ex 3.3
- Ex 3.4
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- Miscellaneous
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.