The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=y \tan x-y^2 \sec x$ is

If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors such that $\overline{\mathrm{a}} \neq \overline{0}$ and $\overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \overline{\mathrm{a}} \times \overline{\mathrm{c}},|\overline{\mathrm{a}}|=|\overline{\mathrm{c}}|=1,|\overline{\mathrm{~b}}|=4$ and $|\overline{\mathrm{b}} \times \overline{\mathrm{c}}|=\sqrt{15}$. If $\overline{\mathrm{b}}-2 \overline{\mathrm{c}}=\lambda \overline{\mathrm{a}}$, then $\lambda$ is

Two friends A and B apply for a job in the same company. The probabilities of A getting selected is $\frac{2}{5}$ and that of B is $\frac{4}{7}$. Then the probability, that one of them is selected, is

If $P_1$ and $P_2$ are perpendicular distances (in units) from point $(2,-1)$ to the pair of lines $2 x^2-5 x y+2 y^2=0$, then the value of $\mathrm{P}_1 \mathrm{P}_2$ is