Example 12 - Find value of tan 13pi/12 - Chapter 3 Class 11 - (x + y) formula

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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Eπ‘₯ample, 12 Find the value of tan 13πœ‹/12 tan 13Ο€/12 Putting Ο€ = 180Β° = tan ( (13 Γ— 180Β°)/12 ) = tan (13 Γ— 15Β° ) = tan (195Β° ) = sin⁑〖195Β°γ€—/π‘π‘œπ‘ β‘γ€–195Β°γ€— = 𝑠𝑖𝑛⁑〖(180Β° + 15Β°)γ€—/γ€–π‘π‘œπ‘  〗⁑〖(180Β° + 15Β°)γ€— We know that sin (180 + ΞΈ ) = – sin ΞΈ & cos (180 + ΞΈ ) = – cos ΞΈ = γ€–βˆ’sin〗⁑〖15Β°γ€—/γ€–βˆ’cos 〗⁑〖15Β°γ€— = 𝑠𝑖𝑛⁑〖15Β°γ€—/γ€–π‘π‘œπ‘  〗⁑〖15Β°γ€— = tan 15Β° = tan (45Β° – 30Β°) Using tan (x - y) = (π‘‘π‘Žπ‘›π‘₯ βˆ’ π‘‘π‘Žπ‘›β‘π‘¦)/(1+π‘‘π‘Žπ‘› π‘₯ π‘‘π‘Žπ‘›β‘π‘¦ ) Here putting x = 45Β° and y = 30Β° = (tan 45"Β°" βˆ’ γ€– tan〗⁑30"Β°" )/(1 + tan 45"Β°" tan⁑30"Β°" ) = (1 βˆ’ 1/√3)/(1 + 1 Γ— 1/√3) = ((√3 βˆ’ 1" " )/√3)/((√3 + 1" " )/√3) = (√3 βˆ’1)/√3 Γ— √3/(√3 + 1) = (√3 βˆ’ 1)/(√3 + 1) Rationalizing the same = (√3 βˆ’ 1)/(√3 + 1) Γ— (√3 βˆ’ 1)/(√3 βˆ’ 1) = (√3 βˆ’ 1)2/(√3 + 1)(√3 βˆ’ 1) Using (a – b )2 = a2 + b2 – 2ab = ((√3)2 +(1)2 βˆ’2" " Γ— √3 Γ— 1)/(√3 + 1)(√3 βˆ’ 1) = (3 + 1 βˆ’ 2√3)/(√(3 )+ 1)(√3 βˆ’ 1) Using (a – b ) (a + b) = a2 – b2 = (4 βˆ’ 2√3)/((√3)2 βˆ’ (1)2) = (4 βˆ’ 2√3)/(3 βˆ’ 1) = (4 βˆ’ 2√3)/2 = (2 (2 βˆ’ √(3 )))/2 = 2 – √3 Hence tan 13Ο€/12 = 2 – √3

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