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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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  • Suresh Kumar's image
    Sir aapne example 12 me tan 195 degree ko sidhe hi 180 15 karke tan 15 degree kyo nahi lay. Sin aur cos ka process karne ki kya zarurat thi.
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