Question

If $\frac{{\cos }^{4}x}{{\cos }^{2}y}+\frac{{\sin }^{4}x}{{\sin }^{2}y}=1$, then $\frac{{\cos }^{4}y}{{\cos }^{2}x}+\frac{{\sin }^{4}y}{{\sin }^{2}x}=?$.

Answer 1

$\frac{co{s}^{4}x}{co{s}^{2}y}+\frac{si{n}^{4}x}{si{n}^{2}y}$

Multiply by ${\sin }^{2}y.co{s}^{2}y$

then write ${\sin }^{2}y=1-{\cos }^{2}y$

${\cos }^{2}x(1-{\cos }^{2}y)={\cos }^{4}x+{\cos }^{2}y({\sin }^{4}x-{\cos }^{2}x)$

$={\cos }^{4}x+{\cos }^{2}y({\sin }^{2}x-{\cos }^{2}x)$

$={\cos }^{4}x{\cos }^{2}y-2{\cos }^{2}x{\cos }^{2}y=0$

$({\cos }^{2}x-{\cos }^{2}y{)}^2=0$

${\cos }^{2}x-{\cos }^{2}y=0$

and ${\sin }^{2}x={\sin }^{2}y$

$\frac{co{s}^{4}x}{co{s}^{2}y}+\frac{si{n}^{4}x}{si{n}^{2}y}={\cos }^{2}y+{\sin }^{2}y=1$