Question

Consider the system of equations
$x{\cos }^{3}y+3x\cos y{\sin }^{2}y=14$
$x{\sin }^{3}y+3x{\cos }^{2}y\sin y=13$

The value of ${\sin }^{2}y+2{\cos }^{2}y$ is

• A. $\frac{4}{5}$
• B. $\frac{9}{5}$
• C. $2$
• D. None of these

Answer 1

The correct option is B $\frac{9}{5}$

$x{\cos }^{3}y+3x\cos y{\sin }^{2}y=14\cdots \left(1\right)$
$x{\sin }^{3}y+3x{\cos }^{2}y\sin y=13\cdots \left(2\right)$
Adding Eqs. $\left(1\right)$ and $\left(2\right)$
$x({\cos }^{3}y+3{\cos }^{2}y\sin y+3\cos y{\sin }^{2}y+{\sin }^{3}y)=27$
$\Rightarrow x(\cos y+\sin y{)}^3=27$
$\Rightarrow x^{1/3}(\cos y+\sin y)=3\cdots \left(3\right)$

Subtracting Eq $\left(2\right)$ from Eq $\left(1\right)$, we get
$x({\cos }^{3}y+3\cos y{\sin }^{2}y-3{\cos }^{2}y\sin y-{\sin }^{3}y)=1$
$\Rightarrow x(\cos y-\sin y{)}^3=1$
$\Rightarrow x^{1/3}(\cos y-\sin y)=1\cdots \left(4\right)$

Dividing Eq $\left(3\right)$ by Eq $\left(4\right)$, we get
$\cos y+\sin y=3\cos y-3\sin y$
$\Rightarrow \tan y=\frac{1}{2}$
$\Rightarrow \sin y=\frac{1}{\sqrt{5}},\cos y=\frac{2}{\sqrt{5}}$
or $\sin y=\frac{-1}{\sqrt{5}},\cos y=\frac{-2}{\sqrt{5}}$

$\therefore {\sin }^{2}y+2{\cos }^{2}y=\frac{1}{5}+\frac{8}{5}=\frac{9}{5}$