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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(cos t + log tan t/2)`, y = a sin t
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Solution
Here x = `a(cos t + log tan t/2)` ....(1)
y = a sin t ....(2)
Differentiating (1) and (2) w.r.t. t, we get
`dx/dt = a [- sin t + 1/tan(t/2) d/dt tan t/2]`
= `a [- sin t + 1/(tan t/2). sec^2 t/2 1/2]`
= `(a cos^2 t)/(sin t)`
`dy/dt` = a cos t
∴ `dy/dx = (dy/dt)/(dx/dt)`
= `(a cos t)/((a cos^2 t)/(sin t))`
= tan t
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