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Question
If x and y are connected parametrically by the equation, without eliminating the parameter, find `dy/dx.`
x = a sec θ, y = b tan θ
Solution
Given, x = a sec θ and y = b tan θ.
Differentiating both sides with respect to θ,
`dx/(d theta) = a sec θ tan θ ` and `dy/(d theta) = b sec^2 θ`
`therefore dy/dx = (dy/(d theta))/(dx/(d theta))`
`= (b sec^2 theta)/(a sec theta tan theta) = (b sec theta)/(a tan theta)`
`= b/a sec theta cot theta`
`= b/a xx 1/(cos theta) * (cos theta)/(sin theta)`
`= b/a xx 1/(sin theta)`
`= b/a cosec theta`
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