Last updated at May 29, 2018 by Teachoo

Transcript

Example 15 Prove that (sin θ − cos θ + 1)/(sin θ + cos θ − 1)=1/(sec θ − tan θ) , using the identity sec2 θ=1+tan2 θ. Taking L.H.S (sinθ − cos θ + 1)/(sin θ + cos θ − 1) Divide the numerator and denominator by cos 𝜃 = (1/(cos θ) (sin θ − cos θ +1))/(1/(cos θ)(sin θ + cos θ − 1)) = (((sin θ)/(cos θ)) − ((cos θ)/(cos θ)) + (1/(cos θ)))/(((sin θ)/(cos θ)) + ((cos θ)/(cos θ)) − (1/(cos θ)) ) = ((tan θ − 1 + sec θ))/((tan θ + 1 − cos θ) ) = ((tan θ + sec θ − 1))/((tan θ − sec θ + 1)) Multiplying both numerator and denominator by (tan 𝜃 – sec 𝜃) = ((tan θ + sec θ − 1) (tan θ − sec θ ))/((tan θ − sec θ + 1) (tan θ − sec θ )) = ({(tanθ + secθ ) − 1}(tan θ − sec θ ))/({(tan θ − sec θ) + 1}(tan θ − sec θ )) = ((tan θ + sec θ)(tan θ − sec θ) − 1 × (tan θ − sec θ))/({(tanθ− secθ + 1)} (tan θ − sec θ)) = ((tan2 θ − sec2 θ) − (tan θ − sec θ ))/({(tanθ − secθ + 1)}(tan θ − sec θ )) = ((−1) − (tan θ − sec θ))/({(tanθ− secθ + 1)}(tan θ − sec θ )) = (−1 −tan θ + sec θ)/({(tanθ− secθ + 1)}(tan θ − sec θ )) = (−(1 + tan θ − sec θ))/({(tanθ− secθ + 1)}(tan θ − sec θ )) = (−1)/( (tan θ − sec θ)) = 1/( −(tan θ − sec θ) ) = 1/(sec θ − tan θ ) = R.H.S Since , L.H.S = R.H.S Hence proved

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Example 6 Important

Example 7 Important

Example 11 Important

Example 15 Important You are here

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Ex 8.1, 10 Important

Ex 8.1, 9 Important

Ex 8.3, 5 Important

Ex 8.1, 11 Important

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Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 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.