Question

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

• A. True
• B. False

Answer 1

The correct option is A True

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

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

$⟹(1-{\sin }^{2}A{)}^2{\sin }^{2}B+{\sin }^{4}A(1-{\sin }^{2}B)=(1-{\sin }^{2}B){\sin }^{2}B$

$⟹{\sin }^{2}B(1+{\sin }^{4}A-2{\sin }^{2}A)+{\sin }^{4}A-{\sin }^{4}A{\sin }^{2}B={\sin }^{2}B-{\sin }^{4}B$

$⟹{\sin }^{2}B+{\sin }^{4}A{\sin }^{2}B-2{\sin }^{2}A{\sin }^{2}B+{\sin }^{4}A-{\sin }^{4}A{\sin }^{2}B-{\sin }^{2}B+{\sin }^{4}B=0$

$⟹{\sin }^{4}A+{\sin }^{4}B-2{\sin }^{2}A{\sin }^{2}B=0$

$⟹({\sin }^{2}A-{\sin }^{2}B{)}^2=0$

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

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

So the given relation is $\text{True}$