Question

If $\frac{{\cos }^{4}A}{{\cos }^{2}B}+\frac{{\sin }^{4}A}{{\sin }^{2}B}=1$, then which of the following is/are correct?

• A. ${\sin }^{4}A+{\sin }^{4}B=2{\sin }^{2}A{\sin }^{2}B$
• B. ${\sin }^{4}A+{\sin }^{4}B=2{\cos }^{2}A{\cos }^{2}B$
• C. $\frac{{\cos }^{4}B}{{\cos }^{2}A}+\frac{{\sin }^{4}B}{{\sin }^{2}A}=2$
• D. $\frac{{\cos }^{4}B}{{\cos }^{2}A}+\frac{{\sin }^{4}B}{{\sin }^{2}A}=1$

Answer 1

Answer: (A), (D)

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

$\frac{{\cos }^{4}A}{{\cos }^{2}B}+\frac{{\sin }^{4}A}{{\sin }^{2}B}=1\Rightarrow {\cos }^{4}A{\sin }^{2}B+{\sin }^{4}A{\cos }^{2}B={\cos }^{2}B{\sin }^{2}B$
$\Rightarrow (1-{\sin }^{2}A{)}^2{\sin }^{2}B+{\sin }^{4}A(1-{\sin }^{2}B)=(1-{\sin }^{2}B){\sin }^{2}B\Rightarrow (1-2{\sin }^{2}A+{\sin }^{4}A){\sin }^{2}B+({\sin }^{4}A(1-{\sin }^{2}B))=(1-{\sin }^{2}B){\sin }^{2}B\Rightarrow {\sin }^{4}B-2{\sin }^{2}A{\sin }^{2}B+{\sin }^{4}A=0\Rightarrow ({\sin }^{2}B-{\sin }^{2}A{)}^2=0\Rightarrow {\sin }^{2}A={\sin }^{2}B\Rightarrow {\cos }^{2}A={\cos }^{2}B$

Now, from the given options,
${\sin }^{4}A+{\sin }^{4}B={\sin }^{4}B+{\sin }^{4}B=2{\sin }^{4}B=2{\sin }^{2}A{\sin }^{2}B$

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


Alternate method:
We know that,
${\sin }^2\theta +{\cos }^2\theta =1$
Given equation:
$\frac{{\cos }^{4}A}{{\cos }^{2}B}+\frac{{\sin }^{4}A}{{\sin }^{2}B}=1$
This will be an identity when,
$A=B=\theta$

Now, from the given options,
Clearly, $\frac{{\cos }^{4}B}{{\cos }^{2}A}+\frac{{\sin }^{4}B}{{\sin }^{2}A}=1$ and ${\sin }^{4}A+{\sin }^{4}B=2{\sin }^{2}A{\sin }^{2}B$ hold true.