Question

If $\alpha +\beta ={90}^o$ and $\alpha =2\beta$, then ${\cos }^2\alpha +{\sin }^2\beta$ equals to

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $\frac{1}{2}$

${\cos }^2\alpha +{\sin }^2\beta$
$=={\cos }^2(90-\beta )+{\sin }^2\beta ={\sin }^2\beta +{\sin }^2\beta =2{\sin }^2\beta \therefore \alpha +\beta ={90}^o$
$\alpha =2\beta$
$\Rightarrow 2\beta +\beta ={90}^o\Rightarrow \beta ={30}^o$
$\Rightarrow 2{\sin }^2\beta =2\times {\sin }^2{30}^o=2\times \frac{1}{4}=\frac{1}{2}$