In $$\triangle \mathrm{ABC}$$, with usual notations, $$\mathrm{m} \angle \mathrm{C}=\frac{\pi}{2}$$, if $$\tan \left(\frac{A}{2}\right)$$ and $$\tan \left(\frac{B}{2}\right)$$ are the roots of the equation $$a_1 x^2+b_1 x+c_1=0\left(a_1 \neq 0\right)$$, then

Two sides of a triangle are $$\sqrt{3}+1$$ and $$\sqrt{3}-1$$ and the included angle is $$60^{\circ}$$, then the difference of the remaining angles is

In a triangle ABC with usual notations a = 2, b = 3, then value of $$\frac{\cos 2 \mathrm{~A}}{\mathrm{a}^2}-\frac{\cos 2 \mathrm{~B}}{\mathrm{~b}^2}$$ is

In any $$\triangle A B C$$, with usual notations, $$c(a \cos B-b \cos A)=$$