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Prove that cos⁑ θ - sin⁑θ  + 1 /cos⁑ θ + sin⁑θ  - 1 = cosec θ + cot θ

This is a question of CBSE Sample Paper - Class 10 - 2017/18.

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Prove cos⁑ - sin ⁑+ 1 / cos + sin - 1 = cosec + cot - Teachoo

Question 27 - CBSE Class 10 Sample Paper for 2018 Boards - Part 2
Question 27 - CBSE Class 10 Sample Paper for 2018 Boards - Part 3

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Transcript

Question 27 Prove that cosβ‘γ€–πœƒ βˆ’ sinβ‘πœƒ + 1γ€—/cosβ‘γ€–πœƒ + sinβ‘πœƒ βˆ’ 1γ€— = cosec πœƒ + cot πœƒ Solving RHS cosec πœƒ + cot πœƒ Converting into cos and sin = 1/(sin πœƒ)+ cosβ‘πœƒ/sinβ‘πœƒ = (1 + cosβ‘πœƒ)/sinβ‘πœƒ Solving LHS cosβ‘γ€–πœƒ βˆ’ sinβ‘πœƒ + 1γ€—/cosβ‘γ€–πœƒ + sinβ‘πœƒ βˆ’ 1γ€— = cosβ‘γ€–πœƒ βˆ’ sinβ‘πœƒ + 1γ€—/((cosβ‘γ€–πœƒ + sinβ‘πœƒ) βˆ’ 1γ€— ) Question 27 Prove that cosβ‘γ€–πœƒ βˆ’ sinβ‘πœƒ + 1γ€—/cosβ‘γ€–πœƒ + sinβ‘πœƒ βˆ’ 1γ€— = cosec πœƒ + cot πœƒ Solving RHS cosec πœƒ + cot πœƒ Converting into cos and sin = 1/(sin πœƒ)+ cosβ‘πœƒ/sinβ‘πœƒ = (1 + cosβ‘πœƒ)/sinβ‘πœƒ Solving LHS cosβ‘γ€–πœƒ βˆ’ sinβ‘πœƒ + 1γ€—/cosβ‘γ€–πœƒ + sinβ‘πœƒ βˆ’ 1γ€— = cosβ‘γ€–πœƒ βˆ’ sinβ‘πœƒ + 1γ€—/((cosβ‘γ€–πœƒ + sinβ‘πœƒ) βˆ’ 1γ€— ) = γ€–(cosγ€—β‘γ€–πœƒ βˆ’ sinβ‘πœƒ) + 1γ€—/((cosβ‘γ€–πœƒ + sinβ‘πœƒ) βˆ’ 1γ€— ) Γ— ((cosβ‘πœƒ+ sinβ‘πœƒ ) + 1)/((cosβ‘πœƒ+ sinβ‘πœƒ ) + 1) = (γ€–(cosγ€—β‘πœƒ + 1) βˆ’ sinβ‘πœƒ)/((cosβ‘γ€–πœƒ + sinβ‘πœƒ) βˆ’ 1γ€— ) Γ— ((cosβ‘πœƒ + 1) + sinβ‘πœƒ)/((cosβ‘πœƒ+ sinβ‘πœƒ ) + 1) Using (a – b) (a + b) = a2 – b2 in numerator and denominator = (γ€–γ€–(cosγ€—β‘πœƒ + 1)γ€—^2 βˆ’ sin^2β‘πœƒ)/((cosβ‘πœƒ+ sinβ‘πœƒ )^2 βˆ’ 1^2 ) = (cos^2β‘πœƒ + 1^2 + 2(1) cosβ‘πœƒ βˆ’ sin^2β‘πœƒ)/(cos^2β‘πœƒ + sin^2β‘πœƒ + 2 cosβ‘πœƒ sinβ‘πœƒ βˆ’ 1) = (cos^2β‘πœƒ +1 + 2 cosβ‘πœƒ βˆ’ sin^2β‘πœƒ)/(cos^2β‘πœƒ + sin^2β‘πœƒ + 2 cosβ‘πœƒ sinβ‘πœƒ βˆ’ 1) Using cos^2β‘πœƒ + sin^2β‘πœƒ = 1 in denominator = (cos^2β‘πœƒ +1 + 2 cosβ‘πœƒ βˆ’ sin^2β‘πœƒ)/(1 + 2 cosβ‘πœƒ sinβ‘πœƒ βˆ’ 1) = (cos^2β‘πœƒ + 1 + 2 cosβ‘πœƒ βˆ’ sin^2β‘πœƒ)/(2 cosβ‘πœƒ sinβ‘πœƒ ) = (cos^2β‘πœƒ + 2 cosβ‘πœƒ + (1 βˆ’ sin^2β‘πœƒ ))/(2 cosβ‘πœƒ sinβ‘πœƒ ) Using cos^2β‘πœƒ=1βˆ’ sin^2β‘πœƒ in numerator = (cos^2β‘πœƒ + 2 cosβ‘πœƒ + cos^2β‘πœƒ)/(2 cosβ‘πœƒ sinβ‘πœƒ ) = (2 cos^2β‘πœƒ + 2 cosβ‘πœƒ )/(2 cosβ‘πœƒ sinβ‘πœƒ ) = (2 cosβ‘πœƒ (cosβ‘πœƒ + 1) )/(2 cosβ‘πœƒ sinβ‘πœƒ ) = (cosβ‘πœƒ + 1)/sinβ‘πœƒ = RHS ∴ LHS = RHS 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.