Examples

Chapter 8 Class 10 Introduction to Trignometry
Serial order wise

### Transcript

Example 12 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. Solving L.H.S (sin⁡θ − cos θ + 1)/(sin θ + cos θ − 1) Dividing the numerator & 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 & 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)) Using (a – b) (a + b) = a2 – b2 a = tan θ , b = sec θ = ((𝐭𝐚𝐧𝟐 𝛉 − 𝐬𝐞𝐜𝟐 𝜽) − (tan θ − sec θ ))/({(tan⁡θ − sec⁡θ + 1)}(tan θ − sec θ )) We know that sec2 θ = 1 + tan2 θ sec2 θ – tan2 θ = 1 tan2 θ – sec2 θ = –1 = ((−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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.