Question

If $\frac{co{s}^{4}x}{co{s}^{2}y}+\frac{si{n}^{4}x}{si{n}^{2}y}=1$ then $\frac{co{s}^{4}y}{co{s}^{2}x}+\frac{si{n}^{4}y}{si{n}^{2}x}=$

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Given,

$\frac{{\cos }^{4}x}{{\cos }^{2}y}+\frac{{\sin }^{4}x}{{\sin }^{2}y}=1$

$\frac{({\cos }^{2}x{)}^2}{{\cos }^{2}y}+\frac{{\sin }^{4}x}{{\sin }^{2}y}=1$

$\frac{(1-{\sin }^{2}x{)}^2}{1-{\sin }^{2}y}+\frac{{\sin }^{4}x}{{\sin }^{2}y}=1$

$\frac{{\sin }^{2}y[1+{\sin }^{4}x-2{\sin }^{2}x]+{\sin }^{4}x(1-{\sin }^{2}y)}{{\sin }^{2}y(1-{\sin }^{2}y)}=1$

Upon further simplification, we get,

${\sin }^{4}x-2{\sin }^{2}x{\sin }^{2}y+{\sin }^{4}y=0$

$({\sin }^{2}x-{\sin }^{2}y)=0$

$\Rightarrow {\sin }^{2}x={\sin }^{2}y$.......(1)

$1-{\cos }^{2}x=1-{\cos }^{2}y$

$\Rightarrow {\cos }^{2}x={\cos }^{2}y$........(2)

Given,

$\frac{{\cos }^{4}y}{{\cos }^{2}x}+\frac{{\sin }^{4}y}{{\sin }^{2}x}$

$=\frac{({\cos }^{2}x{)}^2}{{\cos }^{2}x}+\frac{\left({\sin }^{2}x\right)}{{\sin }^{2}x}$

$={\cos }^{2}x+{\sin }^{2}x=1$

$\therefore \frac{{\cos }^{4}y}{{\cos }^{2}x}+\frac{{\sin }^{4}y}{{\sin }^{2}x}=1$