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Example 15 - Prove that (sin - cos θ + 1)/(sin + cos - 1) - Examples

Example 15 - Chapter 8 Class 10 Introduction to Trignometry - Part 2
Example 15 - Chapter 8 Class 10 Introduction to Trignometry - Part 3

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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 (sin⁡θ − 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 )) = ({(tan⁡θ + sec⁡θ ) − 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))/({(tan⁡θ− sec⁡θ + 1)} (tan θ − sec θ)) = ((tan2 θ − sec2 θ) − (tan θ − sec θ ))/({(tan⁡θ − sec⁡θ + 1)}(tan θ − sec θ )) = ((−1) − (tan θ − sec θ))/({(tan⁡θ− sec⁡θ + 1)}(tan θ − sec θ )) = (−1 −tan θ + sec θ)/({(tan⁡θ− sec⁡θ + 1)}(tan θ − sec θ )) = (−(1 + tan θ − sec θ))/({(tan⁡θ− sec⁡θ + 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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.