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

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

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

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

$onsolving$

$={\sin }^{2}x-{\sin }^2=0$

$(1-{\cos }^{2}x)=\left({\cos }^{2}y\right)$

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

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

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

$=1$