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

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

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

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

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

$On\text{Solving,}$

$\text{si}{\text{n}}^{2}x-{\sin }^{2}y=0\to \left(1\right)$

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

$\left({\cos }^{2}x\right)92=\left({\cos }^{2}y\right)\to \left(2\right)$

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