# Find $$dy\over dx$$ where x = a{cos t + $${1\over 2} log tan^2 {t\over 2}$$} and y = a sin t

## 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

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