### Prove that cosβ‘ θ - sinβ‘θ + 1 /cosβ‘ 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/

Last updated at Sept. 14, 2018 by Teachoo

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/

Transcript

Question 27 Prove that cosβ‘γπ β sinβ‘π + 1γ/cosβ‘γπ + sinβ‘π β 1γ = cosec π + cot π Solving RHS cosec π + cot π Converting into cos and sin = 1/(sin π)+ cosβ‘π/sinβ‘π = (1 + cosβ‘π)/sinβ‘π Solving LHS cosβ‘γπ β sinβ‘π + 1γ/cosβ‘γπ + sinβ‘π β 1γ = cosβ‘γπ β sinβ‘π + 1γ/((cosβ‘γπ + sinβ‘π) β 1γ ) Question 27 Prove that cosβ‘γπ β sinβ‘π + 1γ/cosβ‘γπ + sinβ‘π β 1γ = cosec π + cot π Solving RHS cosec π + cot π Converting into cos and sin = 1/(sin π)+ cosβ‘π/sinβ‘π = (1 + cosβ‘π)/sinβ‘π Solving LHS cosβ‘γπ β sinβ‘π + 1γ/cosβ‘γπ + sinβ‘π β 1γ = cosβ‘γπ β sinβ‘π + 1γ/((cosβ‘γπ + sinβ‘π) β 1γ ) = γ(cosγβ‘γπ β sinβ‘π) + 1γ/((cosβ‘γπ + sinβ‘π) β 1γ ) Γ ((cosβ‘π+ sinβ‘π ) + 1)/((cosβ‘π+ sinβ‘π ) + 1) = (γ(cosγβ‘π + 1) β sinβ‘π)/((cosβ‘γπ + sinβ‘π) β 1γ ) Γ ((cosβ‘π + 1) + sinβ‘π)/((cosβ‘π+ sinβ‘π ) + 1) Using (a β b) (a + b) = a2 β b2 in numerator and denominator = (γγ(cosγβ‘π + 1)γ^2 β sin^2β‘π)/((cosβ‘π+ sinβ‘π )^2 β 1^2 ) = (cos^2β‘π + 1^2 + 2(1) cosβ‘π β sin^2β‘π)/(cos^2β‘π + sin^2β‘π + 2 cosβ‘π sinβ‘π β 1) = (cos^2β‘π +1 + 2 cosβ‘π β sin^2β‘π)/(cos^2β‘π + sin^2β‘π + 2 cosβ‘π sinβ‘π β 1) Using cos^2β‘π + sin^2β‘π = 1 in denominator = (cos^2β‘π +1 + 2 cosβ‘π β sin^2β‘π)/(1 + 2 cosβ‘π sinβ‘π β 1) = (cos^2β‘π + 1 + 2 cosβ‘π β sin^2β‘π)/(2 cosβ‘π sinβ‘π ) = (cos^2β‘π + 2 cosβ‘π + (1 β sin^2β‘π ))/(2 cosβ‘π sinβ‘π ) Using cos^2β‘π=1β sin^2β‘π in numerator = (cos^2β‘π + 2 cosβ‘π + cos^2β‘π)/(2 cosβ‘π sinβ‘π ) = (2 cos^2β‘π + 2 cosβ‘π )/(2 cosβ‘π sinβ‘π ) = (2 cosβ‘π (cosβ‘π + 1) )/(2 cosβ‘π sinβ‘π ) = (cosβ‘π + 1)/sinβ‘π = RHS β΄ LHS = RHS Hence proved

CBSE Class 10 Sample Paper for 2018 Boards

Paper Summary

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Question 26

Question 27 You are here

Question 28

Question 29

Question 30

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.