If $x = \sqrt {{a^{{{\sin }^{ - 1}}t}}} ,y = \sqrt {{a^{{{\cos }^{ - 1}}}}} ,$ show that $\cfrac{{dy}}{{dx}} = - \cfrac{y}{x}.$

Given : $x = \sqrt {{a^{{{\sin }^{ - 1}}t}}}$ and $y = \sqrt {{a^{{{\cos }^{ - 1}}}}} ,$
Differentiating x and y w.r.t. t,we get

$\cfrac{{dx}}{{dt}} = \cfrac{1}{2} \cdot \cfrac{1}{{\sqrt {{a^{{{\sin }^{ - 1}}}}} }} \cdot \cfrac{d}{{dt}}{a^{{{\sin }^{ - 1}}t}} = \cfrac{1}{2} \cdot \cfrac{1}{{\sqrt {{a^{{{\sin }^{ - 1}}t}}} }}{a^{{{\sin }^{ - 1}}t}} \cdot \log a \cdot \cfrac{d}{{{\mathop{\rm d}\nolimits} t}}{\sin ^{ - 1}}t$

$= \cfrac{{\sqrt {{a^{{{\sin }^{ - 1}}t}}} }}{2} \cdot \log a\cfrac{1}{{\sqrt {1 - {t^2}} }}$

$\cfrac{{dy}}{{dt}} = \cfrac{1}{2}\cfrac{1}{{\sqrt {{a^{{{\cos }^{ - 1}}t}}} }} \cdot \cfrac{d}{{dt}}{a^{{{\cos }^{ - 1}}t}} = \cfrac{1}{2} \cdot \cfrac{1}{{\sqrt {{a^{{{\cos }^{ - 1}}}}t} }} \cdot {a^{{{\cos }^{ - 1}}t}} \cdot \log a\cfrac{{ - 1}}{{\sqrt {1 - {t^2}} }}$

$= \cfrac{{\sqrt {{a^{{{\cos }^{ - 1}}}}} }}{2}\log a\cfrac{{ - 1}}{{\sqrt {1 - {t^2}} }}$

therefore,$\cfrac{{dy}}{{dx}} = \cfrac{{\cfrac{{dy}}{{dt}}}}{{\cfrac{{dx}}{{dt}}}}$

$= \cfrac{{\cfrac{{\sqrt {{a^{{{\cos }^{ - 1}}t}}} }}{2} \cdot \log a \cdot \cfrac{{ - 1}}{{\sqrt {1 - {t^2}} }}}}{{\cfrac{{\sqrt {{a^{{{\sin }^{ - 1}}t}}} }}{2} \cdot \log a \cdot \cfrac{1}{{\sqrt {1 - {t^2}} }}}}$

$= - \cfrac{{\sqrt {{a^{{{\cos }^{ - 1}}t}}} }}{{\sqrt {{a^{{{\sin }^{ - 1}}t}}} }} = \cfrac{{ - y}}{x}$