Question

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

Answer 1

$\frac{{\cos }^{4}x}{{\cos }^{2}y}+\frac{{\sin }^{4}x}{{\sin }^{2}y}=1$

$\frac{{\left(1-{\sin }^{2}x\right)}^2}{1-{\sin }^{2}y}+\frac{{\sin }^{4}x}{{\sin }^{2}y}=0$

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

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

${\left({\sin }^{2}x-{\sin }^{2}y\right)}^2=0$

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

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

$Now\frac{{\cos }^{4}y}{{\cos }^{2}x}+\frac{{\sin }^{4}y}{{\sin }^{2}x}$

$\frac{{\cos }^{4}x}{{\cos }^{2}x}+\frac{{\sin }^{4}x}{{\sin }^{2}x}$

${\cos }^{2}x+{\sin }^{2}x=1$