Question

${\sin }^2{\text{2}}^{\text{o}}+{\sin }^2{\text{4}}^{\text{o}}+{\sin }^2{\text{6}}^{\text{o}}+....+{\sin }^2\text{9}{\text{0}}^{\text{o}}$ is equal to

• A. $22$
• B. $23$
• C. $44$
• D. $45$

Answer 1

The correct option is B $23$

Now,

${\sin }^2{\text{2}}^{\text{o}}+{\sin }^2{\text{4}}^{\text{o}}+{\sin }^2{\text{6}}^{\text{o}}+....+{\sin }^2\text{9}{\text{0}}^{\text{o}}$

$=({\sin }^2{\text{2}}^{\text{o}}+{\sin }^2{\text{4}}^{\text{o}}+{\sin }^2{\text{6}}^{\text{o}}+...+{\sin }^2{44}^o)+({\sin }^2{46}^o+...+{\sin }^2{98}^o)+{\sin }^2\text{9}{\text{0}}^{\text{o}}$

$=({\sin }^2{\text{2}}^{\text{o}}+{\sin }^2{\text{4}}^{\text{o}}+{\sin }^2{\text{6}}^{\text{o}}+...+{\sin }^2{44}^o)+({\cos }^2{44}^o+...+{\cos }^2{2}^o)+1$ [ Since ${\sin }^2{x}^o={\cos }^2({90}^o-x^o)$]

$=({\sin }^2{2}^o+{\cos }^2{2}^o)+({\sin }^2{4}^o+{\cos }^2{4}^o)+......+({\sin }^2{44}^o+{\cos }^2{44}^o)+1$

$=1+1+.....+1\left(22\text{th}\right)+1$

$=22+1$

$=23$.