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

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Question 1 Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Misc 12 Find ππ¦/ππ₯, if π¦=12 (1 βcosβ‘π‘ ), π₯=10 (π‘ βsinβ‘π‘ ),βπ/2 " "<π₯< π/2 Here, ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π π¦=12 (1 βcosβ‘π‘ ) π¦=12 β12 cosβ‘π‘ Differentiating π€.π.π‘.π₯. ππ¦/ππ‘ = π(12 β 12 cosβ‘π‘ )/ππ‘ ππ¦/ππ‘ = π(12)/ππ‘ β 12 π(cosβ‘π‘ )/ππ‘ ππ¦/ππ‘ = 0 β 12 (βsinβ‘π‘ ) π π/π π = ππ π¬π’π§β‘π Calculating π π/π π π₯=10 (π‘ βsinβ‘π‘ ) π₯=10π‘ β10 sinβ‘π‘ Differentiating π€.π.π‘.π₯. ππ₯/ππ‘ = π(10 β 10 sinβ‘π‘ )/ππ‘ π/ππ‘ (10t β 10 sint) ππ₯/ππ‘ = π(10 π‘)/ππ‘ β π(10 sinβ‘π‘ )/ππ‘ ππ₯/ππ‘ = 10β10 cosβ‘π‘ π π/π π = ππ(πβπππβ‘π ) Therefore ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = (12 sinβ‘π‘)/10(1 βγ cosγβ‘π‘ ) ππ¦/ππ₯ = (6 πππβ‘π)/(5 (π βγ πππγβ‘π ) ) ππ¦/ππ₯ = (6 . π γπ¬π’π§ γβ‘γπ/πγ πππβ‘γ π/πγ)/(5 (π γπππγ^πβ‘γπ/πγ ) ) ππ¦/ππ₯ = (6 cosβ‘γ π‘/2γ)/(5 γsin γβ‘γπ‘/2γ ) π π/π π = π/π πππ π/π ππ¦/ππ₯ = 6(2 sinβ‘γπ‘/2γ β cosβ‘γ π‘/2γ )/(5 (1 β(1 β 2 sin^2β‘γπ‘/2γ )) ) We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by π/2 sin ΞΈ = 2 πππβ‘γπ½/πγ πππβ‘γπ½/πγ and cos 2ΞΈ = 1 β 2sin2 ΞΈ Replacing ΞΈ by π/2 cos ΞΈ = 1 β 2sin2 π/2 1 β cos ΞΈ = 2sin2 π½/π