Question

The sum of the squares of sin $\theta$ + sin (${90}^{\circ }-\theta$) and cos $\theta$ - cos (${90}^{\circ }-\theta$) is equal to 2.

• A. True
• B. False

Answer 1

The correct option is A

True

${\{sin\theta +sin({90}^{\circ }-\theta \left)\right\}}^2+{\{cos\theta -cos({90}^{\circ }-\theta \left)\right\}}^2=\{sin\theta +cos\theta {\}}^2+\{cos\theta -sin\theta {\}}^2(As,sin({90}^{\circ }-\theta )=cos\theta andcos({90}^{\circ }-\theta )=sin\theta )=si{n}^2\theta +co{s}^2\theta +2sin\theta cos\theta +co{s}^2\theta +si{n}^2\theta -2sin\theta cos\theta =2(si{n}^2\theta +co{s}^2\theta )=2\left(1\right)(As,si{n}^2\theta +co{s}^2\theta =1)=2$