1. Let the common value be $k$:

$\frac{\sin^{-1}x}{a}=\frac{\cos^{-1}x}{b}=\frac{\tan^{-1}y}{c}=k$

So,

$\sin^{-1}x=ak,\quad \cos^{-1}x=bk,\quad \tan^{-1}y=ck$

2. Use the identity:

$\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$

Thus,

$ak+bk=\frac{\pi}{2}$

$k(a+b)=\frac{\pi}{2}$

$k=\frac{\pi}{2(a+b)}$

3. Since $\tan^{-1}y=ck$, we get

$\tan^{-1}y=c\cdot \frac{\pi}{2(a+b)}$

Multiply both sides by $2$:

$\frac{\pi c}{a+b}=2\tan^{-1}y$

Therefore,

$\cos\left(\frac{\pi c}{a+b}\right)=\cos(2\tan^{-1}y)$

4. Use the standard identity:

$\cos(2\theta)=\frac{1-\tan^2\theta}{1+\tan^2\theta}$

Put $\theta=\tan^{-1}y$, so $\tan\theta=y$. Hence,

$\cos(2\tan^{-1}y)=\frac{1-y^2}{1+y^2}$

5. Therefore,

$\boxed{\cos\left(\frac{\pi c}{a+b}\right)=\frac{1-y^2}{1+y^2}}$

So the correct option is D.