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Question
If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.
x = a sec θ, y = b tan θ
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Solution
Here x = a sec θ ...(1)
y = b tan θ ....(2)
Differentiating (1) and (2) w.r.t. θ, we get
`dx/(d θ)` = a sec θ tan θ and `dy/(d θ)` = b sec2 θ
`therefore dy/dx = (dy/(d θ))/(dx/(d θ))`
= `(b sec^2 θ)/(a sec θ tan θ)`
= `(b sec θ)/(a tan θ)`
= `b/a` sec θ cot θ
= `b/a xx 1/(cos θ) xx (cos θ)/(sin θ)`
= `b/a xx 1/(sin θ)`
= `b/a` cosec θ
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