Solution :
We have, x = a{cos t + \({1\over 2} log tan^2 {t\over 2}\)} and y = a sin t
\(\implies\) x = a{cos t + \({1\over 2} \times 2 log tan{t\over 2}\)} and y = a sin t
\(\implies\) x = a{cos t + {\(log tan{t\over 2}\)} and y = a sin t
By using differentiation of parametric functions,
Differentiating with respect to t, we get
\(dx\over dt\) = a{-sin t + \({1\over tan t/2}(sec^2{t\over 2})\times {1\over 2}\)} and \(dy\over dt\) = a cost
\(dx\over dt\) = a{-sin t + \(1\over 2 sin (t/2) cos (t/2)\)} and \(dy\over dt\) = a cost
\(\implies\) \(dx\over dt\) = a{-sin t + \(1\over sint \)} and \(dy\over dt\) = a cost
\(\implies\) \(dx\over dt\) = a{\(-sin^2t + 1\over sint \)} and \(dy\over dt\) = a cost
\(dx\over dt\) = a{\(cos^2t\over sint \)} and \(dy\over dt\) = a cost
\(\therefore\) \(dy\over dx\) = \(dy/dt\over dx/dt\) = \(acost\over {acos^2t\over sint}\)
\(dy\over dx\) = tan t
