### Prove that cos θ - sinθ + 1 /cos θ + sinθ - 1 = cosec θ + cot θ

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Last updated at May 29, 2018 by Teachoo

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Question 27 Prove that cos〖𝜃 − sin𝜃 + 1〗/cos〖𝜃 + sin𝜃 − 1〗 = cosec 𝜃 + cot 𝜃 Solving RHS cosec 𝜃 + cot 𝜃 Converting into cos and sin = 1/(sin 𝜃)+ cos𝜃/sin𝜃 = (1 + cos𝜃)/sin𝜃 Solving LHS cos〖𝜃 − sin𝜃 + 1〗/cos〖𝜃 + sin𝜃 − 1〗 = cos〖𝜃 − sin𝜃 + 1〗/((cos〖𝜃 + sin𝜃) − 1〗 ) Question 27 Prove that cos〖𝜃 − sin𝜃 + 1〗/cos〖𝜃 + sin𝜃 − 1〗 = cosec 𝜃 + cot 𝜃 Solving RHS cosec 𝜃 + cot 𝜃 Converting into cos and sin = 1/(sin 𝜃)+ cos𝜃/sin𝜃 = (1 + cos𝜃)/sin𝜃 Solving LHS cos〖𝜃 − sin𝜃 + 1〗/cos〖𝜃 + sin𝜃 − 1〗 = cos〖𝜃 − sin𝜃 + 1〗/((cos〖𝜃 + sin𝜃) − 1〗 ) = 〖(cos〗〖𝜃 − sin𝜃) + 1〗/((cos〖𝜃 + sin𝜃) − 1〗 ) × ((cos𝜃+ sin𝜃 ) + 1)/((cos𝜃+ sin𝜃 ) + 1) = (〖(cos〗𝜃 + 1) − sin𝜃)/((cos〖𝜃 + sin𝜃) − 1〗 ) × ((cos𝜃 + 1) + sin𝜃)/((cos𝜃+ sin𝜃 ) + 1) Using (a – b) (a + b) = a2 – b2 in numerator and denominator = (〖〖(cos〗𝜃 + 1)〗^2 − sin^2𝜃)/((cos𝜃+ sin𝜃 )^2 − 1^2 ) = (cos^2𝜃 + 1^2 + 2(1) cos𝜃 − sin^2𝜃)/(cos^2𝜃 + sin^2𝜃 + 2 cos𝜃 sin𝜃 − 1) = (cos^2𝜃 +1 + 2 cos𝜃 − sin^2𝜃)/(cos^2𝜃 + sin^2𝜃 + 2 cos𝜃 sin𝜃 − 1) Using cos^2𝜃 + sin^2𝜃 = 1 in denominator = (cos^2𝜃 +1 + 2 cos𝜃 − sin^2𝜃)/(1 + 2 cos𝜃 sin𝜃 − 1) = (cos^2𝜃 + 1 + 2 cos𝜃 − sin^2𝜃)/(2 cos𝜃 sin𝜃 ) = (cos^2𝜃 + 2 cos𝜃 + (1 − sin^2𝜃 ))/(2 cos𝜃 sin𝜃 ) Using cos^2𝜃=1− sin^2𝜃 in numerator = (cos^2𝜃 + 2 cos𝜃 + cos^2𝜃)/(2 cos𝜃 sin𝜃 ) = (2 cos^2𝜃 + 2 cos𝜃 )/(2 cos𝜃 sin𝜃 ) = (2 cos𝜃 (cos𝜃 + 1) )/(2 cos𝜃 sin𝜃 ) = (cos𝜃 + 1)/sin𝜃 = RHS ∴ LHS = RHS Hence proved

Paper Summary

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Question 27 You are here

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Question 30

Class 10

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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.