# Ex 5.6, 8 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 by Teachoo

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Ex 5.6, 8 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ = π(cosβ‘γπ‘+logβ‘γ tanγβ‘γπ‘/2γ γ ) , π¦=π sinβ‘π‘Here ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π π¦ = π sinβ‘π‘ ππ¦/ππ‘ " " = π(π sinβ‘π‘)/ππ‘ ππ¦/ππ‘ " " = π π( sinβ‘π‘)/ππ‘ ππ¦/ππ‘ " " = π cosβ‘π‘ Calculating π π/π π π₯ = π(cosβ‘γπ‘+logβ‘γ tanγβ‘γπ‘/2γ γ ) ππ₯/ππ‘ = π(π(cosβ‘γπ‘+logβ‘γ tan γβ‘γπ‘/2γ γ ))/ππ‘ ππ₯/ππ‘ = a ((π(cosβ‘π‘ ) )/ππ‘ +π(logβ‘(γtan γβ‘γπ‘/2γ ) )/ππ‘) ππ₯/ππ‘ = a (βsinβ‘π‘ +π(logβ‘(tanβ‘γ π‘/2γ ) )/ππ‘) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/tanβ‘γ π‘/2γ . π(γtan γβ‘γπ‘/2γ )/ππ‘) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/tanβ‘γ π‘/2γ . (sec^2β‘γπ‘/2γ ).π(π‘/2)/ππ‘ ) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/tanβ‘γ π‘/2γ . sec^2 (π‘/2). 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +cotβ‘γπ‘/2γ . 1/(cos^2 π‘/2) . 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +γcos γβ‘γπ‘/2γ/sinβ‘γ π‘/2γ . 1/(cos^2 π‘/2) . 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/(sinβ‘γ π‘/2γ cosβ‘γ π‘/2γ ) . 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/(2 sinβ‘γπ‘/2γ cosβ‘γπ‘/2γ )) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/sinβ‘π‘ ) ππ₯/ππ‘ = a ((βsin^2β‘π‘ + 1)/sinβ‘π‘ ) ππ₯/ππ‘ = a ((1 β sin^2β‘π‘)/sinβ‘π‘ ) π π/π π = a (γπππγ^πβ‘π/πππβ‘π ) We know that sin 2ΞΈ =2sin ΞΈ cos ΞΈ Replacing ΞΈ by π‘/2 sin t = 2 sin π‘/2 . cos π‘/2 Therefore, π π/π π " =" (π π/π π)/(π π/π π) ππ¦/ππ₯ = (π cosβ‘π‘)/(γπ cos^2γβ‘π‘/sinβ‘π‘ ) ππ¦/ππ₯ = π cosβ‘π‘ Γ sinβ‘π‘/(π cos^2 π‘) ππ¦/ππ₯ = (π cosβ‘π‘ . sinβ‘π‘)/(π cosβ‘π‘ .cosβ‘π‘ ) ππ¦/ππ₯ = sinβ‘π‘/cosβ‘π‘ π π/π π = πππ§β‘π