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Chapter 3 Class 11 Trigonometric Functions

Serial order wise

Last updated at April 16, 2024 by Teachoo

Example 21 If tanβ‘π₯ = 3/4 , "Ο" < π₯ < 3π/4 , find the value of sin π₯/2 , cos π₯/2 and tan π₯/2 Given that "Ο" < x < ππ /π 180Β° < x < 3/2 Γ 180Β° 180Β° < x < 270Β° Dividing by 2 all sides (180Β°)/2 < π₯/2 < (270Β°)/2 90Β° < π/π < 135Β° So, π₯/2 lies in 2nd quadrant In 2nd quadrant, sin is positive, cos & tan are negative sin π₯/2 is positive, cos π₯/2 and tan π₯/2 are negative Given tan x = 3/4 We know that tan 2x = (2 π‘ππβ‘π₯)/(1 β π‘ππ2π₯) Replacing x with π₯/2 tan (2π₯/2) = (2 π‘ππβ‘(π₯/2))/(1 β π‘ππ2(π₯/2) ) tan x = (2 π‘ππβ‘(π₯/2))/(1 β π‘ππ2(π₯/2) ) tan x = (π πππβ‘(π/π))/(π β ππππ(π/π) ) 3/4 = (2 tanβ‘(π₯/2))/(1 β π‘ππ2(π₯/2) ) 3(1 β tan2 (π₯/2) = 4 Γ 2 tan (π₯/2) 3 Γ 1 β 3 Γ tan2 (π₯/2) = 8 tan (π₯/2) 3 Γ 1 β 3 Γ tan2 (π₯/2) = 8 tan (π₯/2) 3 β 3 tan2 (π₯/2) = 8 tan (π₯/2) 0 = β3 + 3tan2 (π/π) + 8 tan π/π Replacing tan π±/π by a Our equation becomes 0 = β3 + 3a2 + 8a 3a2 + 8a β 3 = 0 3a2 + 9a β a β 3 = 0 3a (a + 3) β 1 (a + 3) = 0 (3a β 1) (a + 3) = 0 Hence So, a = 1/3 or a = β3 Hence, tan π₯/2 = 1/3 or tan π₯/2 = β3 Since π₯/2 lies in IInd quadrant So, tan π/π is negative, β΄ tan π/π = β3 Now, We know that 1 + tan2 x = sec2 x Replacing x with π₯/2 1 + tan2 π₯/2 = sec2 π₯/2 1 + (β3)2 = sec2 π₯/2 1 + 9 = sec2 x/2 1 + 9 = sec2 x/2 10 = sec2 π₯/2 sec2 π₯/2 = 10 sec π/π = Β± βππ Since π₯/2 lie on the llnd Quadrant, cos π₯/2 is negative in the llnd Quadrant β΄ sec π/π is negative in the llnd Quadrant So, sec π/π = ββππ Therefore, cos π/π = (βπ)/βππ Now, We know that sin2x + cos2x = 1 Replacing x with π₯/2 sin2 π₯/2 + cos2 π₯/2 = 1 sin2 π₯/2 = 1 β cos2 π₯/2 Putting cos π₯/2 = β1/β10 sin2 π/π = 1 β ((βπ)/βππ)π sin2 π₯/2 = 1 β 1/10 sin2 π₯/2 = (10 β 1)/10 sin2 π₯/2 = 9/10 sin π₯/2 = Β± β(9/10) sin π₯/2= Β± β9/β10 sin π/π = Β± π/βππ Since π₯/2 lies on the 2nd Quadrant sin π/π is positive in the 2nd Quadrant So, sin π/π = π/βππ Hence, tan π₯/2 = β3 , cos π₯/2 = (β1)/β10 & sin π₯/2 = 3/β10