Question

If $\frac{{\sin }^4\alpha }{{\sin }^2\beta }+\frac{{\cos }^4\alpha }{{\cos }^2\beta }=1$, then

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

Answer 1

Answer: (A), (B)

The correct options are
A ${\sin }^4\alpha +{\sin }^4\beta =2{\sin }^2\alpha {\sin }^2\beta$
B $\frac{{\cos }^4\beta }{{\cos }^2\alpha }+\frac{{\sin }^4\beta }{{\sin }^2\alpha }=1$

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

${\sin }^4\alpha +{\sin }^4\beta =2{\sin }^4\alpha =2{\sin }^2\alpha {\sin }^2\beta$

Also,
$\frac{{\cos }^4\beta }{{\cos }^2\alpha }+\frac{{\sin }^4\beta }{{\sin }^2\alpha }$
$=\frac{{\cos }^2\beta {\cos }^2\alpha }{{\cos }^2\alpha }+\frac{{\sin }^2\beta {\sin }^2\alpha }{{\sin }^2\alpha }$
$={\cos }^2\beta +{\sin }^2\beta =1$