Question

If $\frac{{\cos }^{2}A}{{\cos }^{2}B}+\frac{{\sin }^{4}A}{{\sin }^{2}B}=1$, then show that $\frac{{\cos }^{2}A}{{\cos }^{2}B}+\frac{{\sin }^{4}A}{{\sin }^{2}B}=1$.

Answer 1

We have,

$\frac{{\cos }^{2}A}{{\cos }^{2}B}+\frac{{\sin }^{4}A}{{\sin }^{2}B}=1$

Hence, $\frac{{\cos }^{2}A}{{\cos }^{2}B}+\frac{{\sin }^{4}A}{{\sin }^{2}B}=1$