sin25° + sin210° + sin215° + ...... sin285° + sin290° is equal to?

Answer: Option D

Given, (sin²5° + sin²10° + sin²15° + . . . . . . + sin²85° ) + sin²90°
We know that sinθ = cos(90° - θ )
Therefore sin²85° = cos²(90° - 85°) = cos²5°
Similary sin²60° = cos²(90° - 60°) = cos²40°
And we also know that sin²θ + cos²θ = 1
There are 8 pair in given equation
sin²5° + cos²5° = 1
sin²10° + cos²10° = 1
. . . . . . . . . . . .
. . . . . . . . . . . .
sin²40° + cos²40° = 1
i.e. 8 + sin²45° + sin²90°
⇒ 8 + $$\frac{1}{2}$$ + 1 = $$9\frac{1}{2}$$

Short Trick:
 $$\left(si{n}^2{5}^{\circ }+{\sin }^2{10}^{\circ }+{\sin }^2{15}^{\circ }+.....{\sin }^2{85}^{\circ }\right)+$$         $${\sin }^{2}90$$
$$\begin{array}{rl}& \text{Number of terms}\\ & =\left\{\left(\frac{85-5}{5}\right)+1\right\}\text{ + 1}\\ & \text{ = }\frac{17}{2}\text{ + 1}\\ & \text{Sum of series}=9\frac{1}{2}\end{array}$$