The question goes as follows:
If $\theta$ is the angle between two non-null vectors $A^i$ and $B^i$, show that $sin^2 \theta = \frac{(g_{ij}g_{kl}-g_{ik}g_{jl})A^iB^kA^jB^l}{(g_{ij}A^iA^j)(g_{kl}B^kB^l)}$.
My solution:
We know $cos \ \theta= \frac{g_{ij}A^iB^j}{\sqrt{g_{ij}A^iA^j}\sqrt{g_{ij}B^iB^j}}$
Now, changing the dummy indices in the denominator:
$cos \ \theta= \frac{g_{ij}A^iB^j}{\sqrt{g_{pq}A^pA^q}\sqrt{g_{rs}B^rB^s}}$
$cos^2 \ \theta= \frac{{g_{ij}A^iB^j}{g_{kl}A^kB^l}}{({g_{pq}A^pA^q})({g_{rs}B^rB^s})}$
$1- cos^2 \ \theta= sin^2 \theta= \frac{{g_{pq}A^pA^q}{g_{rs}B^rB^s}-{g_{ij}A^iB^j}{g_{kl}A^kB^l}}{({g_{pq}A^pA^q})({g_{rs}B^rB^s})}$
Changing the dummy indices $p$ and $q$ to $i$ and $j$,
$sin^2 \theta= \frac{{g_{ij}A^iA^j}{g_{kl}B^kB^l}-{g_{ij}A^iB^j}{g_{kl}A^kB^l}}{({g_{ij}A^iA^j})({g_{kl}B^kB^l})}$
Changing $j$ to $k$ in the numerator in the second term, we get:
$sin^2 \theta = \frac{(g_{ij}g_{kl}-g_{ik}g_{jl})A^iB^kA^jB^l}{(g_{ij}A^iA^j)(g_{kl}B^kB^l)}$
I don't know my solution is correct or not. Kindly verify.
Yes, that's fine. I don't think it was necessary to change the dummy indices from $i$ and $j$ to $p$ and $q$ and back again, but if you find it clearer to have all the indices distinct as some point, no harm done.