66The given equation is equivalent to
\dfrac{\tan^{-1}(x+h)-\tan^{-1}x}{h}=\sin^2y+h\times \mathrm{something\ finite}
Taking the limit as h→0, we get
\dfrac{\mathrm d}{\mathrm dx}\tan^{-1}x=\sin^2y
That is \sin^2y=\dfrac{1}{1+x^2}
Hence,
\sin y=\dfrac{1}{\sqrt{1+x^2}}
P.S: By the way, use \sin y to get \sin y and NOT siny to get sin y