If the acute angle between the line $\overline{\mathrm{r}}=\mathrm{i}+2\mathrm{j}+\mathrm{\lambda }(4\mathrm{i}-3\mathrm{k})\text{ and }\mathrm{xy}-\text{ plane is }{\mathrm{\theta }}_1$and acute angle between the planes $\mathrm{x}+2\mathrm{y}=0,2\mathrm{x}+\mathrm{y}=0\text{ is }{\mathrm{\theta }}_2\text{; then }{\cos }^2{\mathrm{\theta }}_1+{\sin }^2{\mathrm{\theta }}_2=$

$\begin{array}{l}{\theta }_1 is the angle between the line \overline{\mathrm{r}}=\mathrm{i}+2\mathrm{j}+\mathrm{\lambda }(4\mathrm{i}-3\mathrm{k})\\ and xy plane\\ \Rightarrow drs of normal to the xy plane\left(0,0,1\right)\\ \cos \left(90°-{\theta }_1\right)=\frac{(4\hat{\mathrm{i}}-3\hat{\mathrm{k}}).\hat{k}}{\sqrt{16+9}}\Rightarrow \sin {\mathrm{\theta }}_1=-3/5\\ \therefore \sin {\mathrm{\theta }}_1=-3/5\Rightarrow \cos {\mathrm{\theta }}_1=-4/5,\\ \\ acute angle between the planes x+2y=0 and 2x+y=0 is{\theta }_2\\ \cos {\mathrm{\theta }}_2=\frac{1\left(2\right)+2\left(1\right)}{\sqrt{1+4}\sqrt{4+1}}=\frac{4}{5}\\ {\cos }^2{\mathrm{\theta }}_1+{\sin }^2{\mathrm{\theta }}_2=\frac{16}{25}+\frac{9}{25}=1\end{array}$