If $$(\sin^{-1} x)^2 - (\cos^{-1} x)^2 = a$$; $$0 < x < 1$$, $$a \neq 0$$, then the value of $$2x^2 - 1$$ is

We are given the relation $$\left(\sin^{-1}x\right)^2-\left(\cos^{-1}x\right)^2=a$$ with $$0<x<1$$ and $$a\neq0$$.

For angles in the interval $$(0,\tfrac{\pi}{2})$$ we always have the elementary identity $$\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x.$$ Using this fact lets us re-express the second square in terms of the first.

Put $$\theta=\sin^{-1}x.$$ Then $$\cos^{-1}x=\frac{\pi}{2}-\theta.$$ Substituting these into the given equation yields

$$\theta^{2}-\left(\frac{\pi}{2}-\theta\right)^{2}=a.$$

Now we expand the square on the right side. First write the binomial square formula $$\left(A-B\right)^{2}=A^{2}-2AB+B^{2}.$$ Applying it here with $$A=\frac{\pi}{2}$$ and $$B=\theta,$$ we find

$$\left(\frac{\pi}{2}-\theta\right)^{2}=\left(\frac{\pi}{2}\right)^{2}-2\left(\frac{\pi}{2}\right)\theta+\theta^{2}=\frac{\pi^{2}}{4}-\pi\theta+\theta^{2}.$$

Substituting this explicit expansion back gives

$$\theta^{2}-\left[\frac{\pi^{2}}{4}-\pi\theta+\theta^{2}\right]=a.$$

The two $$\theta^{2}$$ terms cancel because one is positive and the other is negative:

$$\theta^{2}-\frac{\pi^{2}}{4}+\pi\theta-\theta^{2}=a.$$

Simplifying, we keep only the remaining terms:

$$\pi\theta-\frac{\pi^{2}}{4}=a.$$

Now solve for $$\theta$$. First add $$\dfrac{\pi^{2}}{4}$$ to both sides:

$$\pi\theta=a+\frac{\pi^{2}}{4}.$$

Next divide by $$\pi$$:

$$\theta=\frac{a}{\pi}+\frac{\pi}{4}.$$

But recall that $$\theta=\sin^{-1}x,$$ so

$$x=\sin\theta=\sin\!\left(\frac{a}{\pi}+\frac{\pi}{4}\right).$$

The quantity we must evaluate is $$2x^{2}-1.$$ Because $$x=\sin\alpha$$ where $$\alpha=\dfrac{a}{\pi}+\dfrac{\pi}{4},$$ it is convenient to rewrite $$2\sin^{2}\alpha-1$$ using the standard double-angle identity for cosine. First state the identity:

$$\cos2\alpha=1-2\sin^{2}\alpha \quad\Longrightarrow\quad 2\sin^{2}\alpha-1=-\cos2\alpha.$$

We substitute $$\alpha=\dfrac{a}{\pi}+\dfrac{\pi}{4}:$$

$$2x^{2}-1=2\sin^{2}\alpha-1=-\cos\!\bigl(2\alpha\bigr)=-\cos\!\left(2\left(\frac{a}{\pi}+\frac{\pi}{4}\right)\right).$$

Compute the argument of the cosine explicitly:

$$2\left(\frac{a}{\pi}+\frac{\pi}{4}\right)=\frac{2a}{\pi}+\frac{\pi}{2}.$$

Hence

$$2x^{2}-1=-\cos\!\left(\frac{2a}{\pi}+\frac{\pi}{2}\right).$$

Now use the phase-shift identity $$\cos\left(\theta+\frac{\pi}{2}\right)=-\sin\theta.$$ Setting $$\theta=\dfrac{2a}{\pi},$$ we obtain

$$\cos\!\left(\frac{2a}{\pi}+\frac{\pi}{2}\right)=-\sin\!\left(\frac{2a}{\pi}\right).$$

Therefore

$$-\cos\!\left(\frac{2a}{\pi}+\frac{\pi}{2}\right)=\sin\!\left(\frac{2a}{\pi}\right).$$

Substituting this result back, we finally get

$$2x^{2}-1=\sin\!\left(\frac{2a}{\pi}\right).$$

Among the given options, this matches Option B: $$\sin\!\left(\dfrac{2a}{\pi}\right).$$

Hence, the correct answer is Option B.