Examples
Example 2
Example 3
Example 4 Important
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11 Important
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16
Example 17 Important
Example 18
Example 19
Example 20 Important
Example 21
Example 22
Example 23
Example 24 Important
Example 25
Example 26 (i)
Example 26 (ii) Important
Example 26 (iii) Important
Example 26 (iv)
Example 27 Important
Example 28
Example 29 Important
Example 30 Important
Example 31
Example 32
Example 33 Important You are here
Example 34 Important
Example 35
Example 36 Important
Example 37
Example 38 Important
Example 39 (i)
Example 39 (ii) Important
Example 40 (i)
Example 40 (ii) Important
Example 40 (iii) Important
Example 41
Example 42 Important
Example 43
Question 1
Question 2 Important
Question 3
Question 4 Important
Question 5
Question 6 Important
Last updated at Dec. 16, 2024 by Teachoo
Example 33 Find ππ¦/ππ₯ , if π₯ = π (π+sinβ‘π), π¦ = π (1 β cosβ‘π)Here ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating π π/π π½ π¦ = π (1 β cosβ‘π) ππ¦/ππ = π(π (1 β cosβ‘π))/ππ ππ¦/ππ = π (0 β(βsinβ‘π )) ππ¦/ππ = π (πππβ‘π½ ) Calculating π π/π π½ π₯ = π (π+π ππβ‘π) ππ₯/ππ = π(π(π + π ππβ‘π))/ππ ππ₯/ππ = π (ππ/ππ+π(sinβ‘π )/(ππ )) ππ₯/ππ = π (π+ππ¨π¬β‘π½ ) Therefore ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = π" " (sinβ‘π )/π" " (1 +γ cosγβ‘π ) ππ¦/ππ₯ = sinβ‘π/(1 +γ cosγβ‘π ) ππ¦/ππ₯ = (γπ πππγβ‘γ π½/πγ .γ πππγβ‘γ π½/πγ)/(1 + γπ γπππγ^π γβ‘γπ½/πγ β π) ππ¦/ππ₯ = (sinβ‘γ π/2γ )/γcos γβ‘γπ/2γ π π/π π = πππ§β‘γπ½/πγ Rough 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 = 2cos2 ΞΈ β 1 Replacing ΞΈ by π/2 cos ΞΈ = 2cos2 π½/π β 1