Check sibling questions

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.

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

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

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.