Tangents drawn from (b,a) to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ makes angles $\theta_1$ $\theta_2$ with the x-axis. If $\tan\theta_1$$\tan\theta_2$ =2 then $b^2-a^2=....$
The equation of the tangent will be $y=mx+(a−mb)$ and since the condition of tangency is $(a-mb)^2=a^2m^2-b^2$ we get after simplify
$$m^2(b^2-a^2)-2abm+a^2+b^2=0$$
How can I find $(b^2-a^2)$ with a calculator or any easier way than the quadratic formula.
Knowing that $m_1m_2=2$
Converting a comment to an answer, as requested ...
If all you know is that $m_1m_2=2$ (where $m_1$ and $m_2$ are the roots of the quadratic), then you can't find a specific value for $b^2−a^2$.
By Vieta's Formulas, we know that $$\frac{a^2+b^2}{b^2−a^2}=m_1m_2=2$$ From this, we can deduce $b^2=3a^2$, so that $b^2-a^2=2a^2$, but that's as far as we can go.