Question 27 - CBSE Class 10 Sample Paper for 2018 Boards
Last updated at Sept. 14, 2018 by Teachoo

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.

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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

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