sin25° + sin26° + ............. sin284° + sin285° = ?

Answer: Option B

$$\begin{array}{rl}& {\sin }^2{5}^{\circ }+{\sin }^2{6}^{\circ }+.....{\sin }^{2}84^{\circ }+{\sin }^{2}85^{\circ }\\ & \text{Number of terms}\\ & =\frac{85-5}{1}+1\\ & =80+1\\ & =81\end{array}$$
  $$=\left({\sin }^2{5}^{\circ }+{\sin }^2{85}^{\circ }\right)+$$     $$\left({\sin }^2{6}^{\circ }+{\sin }^2{84}^{\circ }\right)+$$     . . . . . upto 40 pairs $$+$$ middle term
$$\begin{array}{rl}& =\text{40}+{\sin }^{2}45^{\circ }\\ & =40+\frac{1}{2}\\ & =40\frac{1}{2}\end{array}$$

Alternate:
In case,
when series is in the form of sin²θ or cos²θ,
then sum of series will always be half of number of terms.
$$=\frac{81}{2}=40\frac{1}{2}$$