The equation of the tangent to the curve given by $x=a{\sin }^{3}t,y={\mathrm{bcos}}^{3}t\text{ at a point where }t=\frac{\pi }{2}$ is

$x=a{\sin }^{3}t\Rightarrow \frac{dx}{dt}=3a{\sin }^{2}t\cdot \cos t$
and $y=b{\cos }^{3}t\Rightarrow \frac{dy}{dt}=3b{\cos }^{2}t\cdot (-\sin t)$
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-3b{\cos }^{2}t\cdot \sin t}{3a{\sin }^{2}t\cdot \cos t}=-\frac{b}{a}\cot t$
$x=\frac{\pi }{2};m=\frac{dy}{dx}=0,x=a{\sin }^3\left(\frac{\pi }{2}\right)=a$
and $y=b{\cos }^3\left(\frac{\pi }{2}\right)=0$
$\therefore \text{ equation of tangent }y-0=0(x-0)\Rightarrow y=0$