Question

The slope of the tangent to
x = a sint, y = a {cost + log (tan)}. At
the point tis​

Answer 1

To find the slope of the tangent at the point (x,y) = (asint, a[cost + log(tant)]

We know that slope of a line can be written as :

$\sf \: m = \dfrac{dy}{dx}$

Multiplying and dividing by dt,

$\longrightarrow \sf \: m = \dfrac{dy}{dx} \: \times \dfrac{dt}{dt} \\ \\ \longrightarrow \sf \: m = \dfrac{dy}{dt} \: \times \dfrac{dt}{dx}$

Thus,

$\sf \: \dfrac{dy}{dt} \implies \: \dfrac{d [a \big(cos \: t \: + log(tan \: t) \big) ] }{dt} \\ \\ \sf \: \dfrac{dy}{dt} \implies a \dfrac{d(cos \: t)}{dt} + a \dfrac{d(log(tan \: t))}{dt} \\ \\ \sf \: \dfrac{dy}{dt} \implies - asin(t) \: + acot(t) \times \dfrac{d(tan \: t)}{dt} \\ \\ \sf \: \dfrac{dy}{dt} \implies - a sin \: t + a\dfrac{cos \: t}{sin \: t} \times {sec}^{2} t \\ \\ \sf \: \dfrac{dy}{dt} \implies \: asec \: t \: cosec \: t - asin \: t - - - - - (1)$

Also,

$\sf \: \dfrac{dx}{dt} \implies \: \dfrac{d(a sin \: t)}{dt} \\ \\ \sf \: \dfrac{dx}{dt} \implies \: acos \: t - - - - - (2)$

From (1) and (2),

$\sf \: m =a (sec \: t \: cosec \: t - \: sin \: t )\times \dfrac{1}{acos \: t} \\ \\ \longrightarrow \boxed{ \boxed{\sf \: m = sec \: t(sec \: t \: cosec \: t - sin \: t)}}$