How is the time period of a pendulum indepedent of the amplitude provided?The time period of the pendulum remains the same if released from a...

It is not true.  The period of a pendulum is always
dependent on the amplitude, but is approximately true for small amplitudes where
sin(theta) is almost theta.

In these equations theta must
be given in radians, and for this to work theta must be much less than 1
radian.

The force on the pendulum is -mg
sin(theta)

s = len*theta

ds/dt
= len * d(theta)/dt

d^2(s)/dt^2 = len * d^2(theta)/dt^2
so

F = ma = m * len * d^2(theta)/dt^2 and our differential
equation is

m (len * d^2(theta)/dt^2) = -mg sin(theta) 
solving we get

d^2(theta)/dt^2 + g/l sin(theta) =
0

This equation cannot be solved in elementary functions,
but if we use sin(theta) is approximately theta

the
differential equation

d^2(theta)/dt^2 + g/l (theta) =
0

gives a solution

theta(t) =
theta(0) cos(sqrt(g/l) t)

The period of such an equation is
2pi/(sqrt(g/l)) or

T = 2pi sqrt(l/g)  where l is the length
of the pendulum and g is the acceleration of gravity and as an approximation this does
not depend on theta(0) the inital angle.