If $$\frac{\sin^{-1}x}{a} = \frac{\cos^{-1}x}{b} = \frac{\tan^{-1}c}{c}$$; $$0 < x < 1$$, then the value of $$\cos\left(\frac{\pi c}{a+b}\right)$$ is:

Let $$\frac{\sin^{-1}x}{a} = \frac{\cos^{-1}x}{b} = \frac{\tan^{-1}y}{c} = k$$ (say), where $$0 < x < 1$$.

Then $$\sin^{-1}x = ak$$ and $$\cos^{-1}x = bk$$. Since $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$, we get $$ak + bk = \frac{\pi}{2}$$, so $$k = \frac{\pi}{2(a+b)}$$.

Also $$\tan^{-1}y = ck$$, so $$\frac{\pi c}{a+b} = 2ck = 2\tan^{-1}y$$.

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

Using the identity $$\cos(2\tan^{-1}y) = \frac{1 - y^2}{1 + y^2}$$ (which follows from letting $$\theta = \tan^{-1}y$$ so that $$\cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1-y^2}{1+y^2}$$), we get $$\cos\left(\frac{\pi c}{a+b}\right) = \frac{1 - y^2}{1 + y^2}$$.