Kwenze0l

2022-09-06

Let ${\stackrel{\to }{V}}_{1}$ and ${\stackrel{\to }{V}}_{2}$ are two vectors such that ${\stackrel{\to }{V}}_{1}=2\left(\mathrm{sin}\alpha +\mathrm{cos}\alpha \right)\stackrel{^}{i}+\stackrel{^}{j}$ and ${\stackrel{\to }{V}}_{2}=\mathrm{sin}\beta \phantom{\rule{thickmathspace}{0ex}}\stackrel{^}{i}+\mathrm{cos}\beta \phantom{\rule{thickmathspace}{0ex}}\stackrel{^}{j}$, where $\alpha$ and $\beta$ satisfy the relation $2\left(\mathrm{sin}\alpha \phantom{\rule{thickmathspace}{0ex}}+\mathrm{cos}\alpha \right)\mathrm{sin}\beta =3-\mathrm{cos}\beta ,$ Find Value of $\left(3{\mathrm{tan}}^{2}\alpha \phantom{\rule{thickmathspace}{0ex}}+4{\mathrm{tan}}^{2}\beta \right)$

Quinn Alvarez

Doesn't really have anything to do with vectors.
Note that the maximum value of $a\mathrm{sin}\theta +b\mathrm{cos}\theta$ is $\sqrt{{a}^{2}+{b}^{2}}$, which occurs when $\mathrm{tan}\theta =\frac{a}{b}$. Here, we've been given that:
$2\left(\mathrm{sin}\alpha +\mathrm{cos}\alpha \right)\mathrm{sin}\beta +\mathrm{cos}\beta =3$
Now, maximum value of LHS is
$\sqrt{4\left(\mathrm{sin}\alpha +\mathrm{cos}\alpha {\right)}^{2}+1}=\sqrt{5+4\mathrm{sin}2\alpha }$
Notice that $\sqrt{5+4\mathrm{sin}2\alpha }\le 3$, reaching maximum value at $\mathrm{sin}2\alpha =1$ Thus, maximum value of LHS, is 3, which is reached when $\mathrm{sin}2\alpha =1$ and $\mathrm{tan}\beta =2\left(\mathrm{sin}\alpha +\mathrm{cos}\alpha \right)$. Now, $\mathrm{sin}2\alpha =1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\mathrm{tan}}^{2}\alpha =1$, and ${\mathrm{tan}}^{2}\beta =4\left(1+\mathrm{sin}2\alpha \right)=8$. Thus, $3{\mathrm{tan}}^{2}\alpha +4{\mathrm{tan}}^{2}\beta =3\cdot 1+4\cdot 8=35$

Do you have a similar question?