Prove that cos⁑ θ - sinβ‘θ  + 1 /cos⁑ θ + sinβ‘θ  - 1 = cosec θ + cot θ

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

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/

Slide93.JPG

Slide94.JPG
Slide95.JPG

  1. Class 10
  2. Sample Papers, Previous Year Papers and Other Questions
Ask Download

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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.