Question

If $\frac{{\cos }^{4}A}{{\cos }^{2}B}+\frac{{\sin }^{4}A}{{\sin }^{2}B}=1$, then $\frac{{\cos }^{4}B}{{\cos }^{2}A}+\frac{{\sin }^{4}B}{{\sin }^{2}A}$ is equal to

Answer 1

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

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

$\frac{{\cos }^{2}A({\cos }^{2}A-{\cos }^{2}B)}{{\cos }^{2}B}={\sin }^{2}A\frac{({\sin }^{2}B-{\sin }^{2}A)}{{\sin }^{2}B}$

$\frac{{\cos }^{2}A}{{\cos }^{2}B}({\cos }^{2}A-{\cos }^{2}B)=\frac{{\sin }^{2}A}{{\sin }^{2}B}[(1-{\cos }^{2}B)-(1-{\cos }^{2}A)]$

$\frac{{\cos }^{2}A}{{\cos }^{2}B}({\cos }^{2}A-{\cos }^{2}B)=\frac{{\sin }^{2}A}{{\sin }^{2}B}({\cos }^{2}A-{\cos }^{2}B)$

$({\cos }^{2}A-{\cos }^{2}B)(\frac{{\cos }^{2}A}{{\cos }^{2}B}-\frac{{\sin }^{2}A}{{\sin }^{2}B})=0$

When ${\cos }^{2}A-{\cos }^{2}B=0$, we have
${\cos }^{2}A={\cos }^{2}B$ (i)

When $\frac{{\cos }^{2}A}{{\cos }^{2}B}-\frac{{\sin }^{2}A}{{\sin }^{2}B}=0$, we have

${\cos }^{2}A{\sin }^{2}B={\sin }^{2}A{\cos }^{2}B$

${\cos }^{2}A(1-{\cos }^{2}B)=(1-{\cos }^{2}A){\cos }^{2}B$

${\cos }^{2}A-{\cos }^{2}A{\cos }^{2}B={\cos }^{2}B-{\cos }^{2}A{\cos }^{2}B$

${\cos }^{2}A={\cos }^{2}B$ (ii)

Thus, in both the cases, ${\cos }^{2}A={\cos }^{2}B$.

$\therefore$ $1-{\sin }^{2}A=1-{\sin }^{2}B$ or ${\sin }^{2}A={\sin }^{2}B$ (iii)

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

Ans: 1