$\sin x = \frac{{2t}}{{1 + {t^2}}},\tan y = \frac{{2t}}{{1 - {t^2}}}$

……(i)
and $\tan y = \frac{{2t}}{{1 - {t^2}}}$ …….(ii)
therefore,$\frac{d}{{dx}}\sin x \cdot \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {\frac{{2t}}{{1 + {t^2}}}} \right)$

$\Rightarrow$ $\cos x\frac{{dx}}{{dt}} = \frac{{\left( {1 + {t^2}} \right) \cdot \frac{d}{{dt}}(2t) - (2t) \cdot \frac{d}{{dt}}\left( {1 + {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}$

$= \frac{{2\left( {1 + {t^2}} \right) - 2t \cdot 2t}}{{{{\left( {1 + {t^2}} \right)}^2}}} = \frac{{2 + 2{t^2} - 4{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}$

$\Rightarrow$ $\frac{{dx}}{{dt}} = \frac{{2\left( {1 - {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}} \cdot \frac{1}{{\cos x}}$

$\Rightarrow$ $\frac{{dx}}{{dt}} = \frac{{2\left( {1 - {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}} \cdot \frac{1}{{\sqrt {1 - {{\sin }^2}x} }} = \frac{{2\left( {1 - {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}} \cdot \frac{1}{{\sqrt {1 - {{\left( {\frac{{2t}}{{1 + {t^2}}}} \right)}^2}} }}$

$\Rightarrow$ $\frac{{dx}}{{dt}} = \frac{{2\left( {1 - {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}} \cdot \frac{{\left( {1 + {t^2}} \right)}}{{\left( {1 - {t^2}} \right)}} = \frac{2}{{1 + {t^2}}}$ …….(iii)

Also, $\frac{d}{{dy}}\tan y \cdot \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\frac{{2t}}{{1 - {t^2}}}} \right)$

${\sec ^2}y\frac{{dy}}{{dt}} = \frac{{\left( {1 - {t^2}} \right)\frac{d}{{dt}} \cdot (2t) - 2t \cdot \frac{d}{{dt}}\left( {1 - {t^2}} \right)}}{{{{\left( {1 - {t^2}} \right)}^2}}}$

$\frac{{dy}}{{dt}} = \frac{{2 - 2{t^2} + 4{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}} \cdot \frac{1}{{{{\sec }^2}y}}$

$= \frac{{2\left( {1 + {t^2}} \right)}}{{{{\left( {1 - {t^2}} \right)}^2}}} \cdot \frac{1}{{\left( {1 + {{\tan }^2}y} \right)}} = \frac{{2\left( {1 + {t^2}} \right)}}{{{{\left( {1 - {t^2}} \right)}^2}}} \cdot \frac{1}{{1 + \frac{{4{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}}}}$

$= \frac{{2\left( {1 + {t^2}} \right)}}{{{{\left( {1 - {t^2}} \right)}^2}}} \cdot \frac{{{{\left( {1 - {t^2}} \right)}^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}} = \frac{2}{{1 + {t^2}}}$ …….(iv)

therefore,$\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{2/1 + {t^2}}}{{2/1 + {t^2}}} = 1$ [from Eqs. (iii) and (iv)]

$\Rightarrow \frac{dy}{dx} = 1$