Find dy/dx in terms of t for the curve given by the parametric equations x = tan(t) , y = sec(t) for -pi/2<t<pi/2.
We know that dy/dx = (dy/dt) * (dt/dx). Differentiating each of the equations with respect to t gives. dy/dt = sec(t) tan(t) and dx/dt = sec2(t). Since dt/dx = 1 / (dx/dt) we have that dt/dx = 1/(sec2(t)) = cos2(t). Substituting back into the first equation gives dy/dx = cos2(t) sec(t) tan(t) . Using the following identities. sec(t) = 1/cos(t) and tan(t) = sin(t)/cos(t) .dy/dx = sin(t)*(cos2(t)/cos2(t))= sin(t). So the final answer is dy/dx = sin(t)
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