Question

If $\alpha +\beta ={90}^0$ 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 $\frac{1}{2}$

$\alpha +\beta ={90}^0$ ...(i)

$\alpha -2\beta =0$ ...(ii)

Subtracting ii from i we get

$3\beta ={90}^0$

$\beta ={30}^0$

Therefore

${\cos }^2\alpha +{\sin }^2\beta$

$={\cos }^2({90}^0-\beta )+{\sin }^2\beta$ ...from(i)

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

$=2{\sin }^2\beta$

Substituting the value of $\beta$ we get

$2{\sin }^2\beta =2\left(\frac{1}{2}\right){}^2$

$=\frac{1}{2}$

Hence the answer is A