Last updated at May 29, 2018 by Teachoo

Transcript

Example, 28 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 < 3π/2 i.e.180Β°< x < 3/2 Γ 180Β° i.e. 180Β° < x < 270Β° Dividing by 2 all sides (180Β°)/2 < π₯/2 < (270Β°)/2 90Β° < π₯/2 < 135Β° So, π₯/2 lies in IInd quadrant In IInd quadrant, sin is positive, cos & tan are negative β sin π₯/2 Positive and cos π₯/2 and tan π₯/2 negative Given tan x = 3/4 tan x = (2 tanβ‘(π₯/2))/(1 β π‘ππ2(π₯/2) ) 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 β 3 tan2 (π₯/2) = 8 tan (π₯/2) 0 = β 3 + 3tan2 (π₯/2) + 8 tan π₯/2 Replacing tan π₯/2 by a 0 = β 3 + 3a2 + 8a β 3 + 3a2 + 8a = 0 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 3a β 1 = 0 3a = 1 a = 1/3 So, a = 1/3 or a = β 3 Hence, tan π₯/2 = 1/3 or tan π₯/2 = β 3 Since, π₯/2 lies in IInd quadrant tan π₯/2 is negative, β΄ tan π₯/2 = β 3 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 10 = sec2 π₯/2 sec2 π₯/2 = 10 sec π₯/2 = Β± β10 Since π₯/2 lie on the llnd Quadrant, cos π₯/2 is negative in the llnd Quadrant sec π₯/2 is negative in the llnd Quadrant So sec π₯/2 = ββ10 sec π₯/2 = ββ10 β 1/cosβ‘γ π₯/2γ = ββ10 β 1/(ββ10) = cos π₯/2 β cos π₯/2 = β1/β10 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₯/2 = 1 β ("β" 1/β10)2 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) = Β± β9/β10 = Β± 3/β10 So, sin π₯/2 = Β± 3/β10 Since π₯/2 lie on the llnd Quadrant , sin π₯/2 is positive in the llnd Quadrant So sin π₯/2 = 3/β10 Hence, tan π₯/2 = - 3 , cos π₯/2 = β1/β10 & sin π₯/2 = 3/β10

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Ex 3.3, 4 Important

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Example 24 Important

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Example 27 Important

Example 28 Important You are here

Misc 4 Important

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Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.