Question

Value of $\frac{{\sin }^2{20}^o+{\cos }^4{20}^o}{{\sin }^4{20}^o+{\cos }^2{20}^o}$ is

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is A $1$

$\frac{{\sin }^2{20}^o+{\cos }^4{20}^o}{{\sin }^4{20}^o+{\cos }^2{20}^o}$
$=\frac{{\sin }^2{20}^0+{\cos }^2{20}^0.{\cos }^2{20}^0}{{\sin }^2{20}^0.{\sin }^2{20}^0+{\cos }^2{20}^0}$
$=\frac{{\sin }^2{20}^0+{\cos }^2{20}^0(1-{\sin }^2{20}^0)}{{\sin }^2{20}^0(1-{\cos }^2{20}^0)+{\cos }^2{20}^0}$
$=\frac{{\sin }^2{20}^0+{\cos }^2{20}^0-{\sin }^2{20}^0{\cos }^2{20}^0}{{\sin }^2{20}^0-{\sin }^2{20}^0{\cos }^2{20}^0+{\cos }^2{20}^0}$
$=\frac{1-{\sin }^2{20}^0{\cos }^2{20}^0}{{\sin }^2{20}^0-{\sin }^2{20}^0{\cos }^2{20}^0+{\cos }^2{20}^0}$
$=\frac{1-{\sin }^2{20}^0{\cos }^2{20}^0}{1-{\sin }^2{20}^0{\cos }^2{20}^0}=1$