Last updated at Dec. 8, 2016 by Teachoo

Transcript

Eπ₯ample, 12 Find the value of tan 13π/12 tan 13Ο/12 Putting Ο = 180Β° = tan ( (13 Γ 180Β°)/12 ) = tan (13 Γ 15Β° ) = tan (195Β° ) = sinβ‘γ195Β°γ/πππ β‘γ195Β°γ = π ππβ‘γ(180Β° + 15Β°)γ/γπππ γβ‘γ(180Β° + 15Β°)γ We know that sin (180 + ΞΈ ) = β sin ΞΈ & cos (180 + ΞΈ ) = β cos ΞΈ = γβsinγβ‘γ15Β°γ/γβcos γβ‘γ15Β°γ = π ππβ‘γ15Β°γ/γπππ γβ‘γ15Β°γ = tan 15Β° = tan (45Β° β 30Β°) Using tan (x - y) = (π‘πππ₯ β π‘ππβ‘π¦)/(1+π‘ππ π₯ π‘ππβ‘π¦ ) Here putting x = 45Β° and y = 30Β° = (tan 45"Β°" β γ tanγβ‘30"Β°" )/(1 + tan 45"Β°" tanβ‘30"Β°" ) = (1 β 1/β3)/(1 + 1 Γ 1/β3) = ((β3 β 1" " )/β3)/((β3 + 1" " )/β3) = (β3 β1)/β3 Γ β3/(β3 + 1) = (β3 β 1)/(β3 + 1) Rationalizing the same = (β3 β 1)/(β3 + 1) Γ (β3 β 1)/(β3 β 1) = (β3 β 1)2/(β3 + 1)(β3 β 1) Using (a β b )2 = a2 + b2 β 2ab = ((β3)2 +(1)2 β2" " Γ β3 Γ 1)/(β3 + 1)(β3 β 1) = (3 + 1 β 2β3)/(β(3 )+ 1)(β3 β 1) Using (a β b ) (a + b) = a2 β b2 = (4 β 2β3)/((β3)2 β (1)2) = (4 β 2β3)/(3 β 1) = (4 β 2β3)/2 = (2 (2 β β(3 )))/2 = 2 β β3 Hence tan 13Ο/12 = 2 β β3

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Chapter 3 Class 11 Trigonometric Functions

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.