Question

Assertion (A) : ${\sin }^2\theta +{\sin }^2(\theta +{60}^0)+{\sin }^2(\theta -{60}^0)=\frac{3}{2}$
Reason (R): $\cos \alpha +\cos ({120}^0+\alpha )+\cos ({120}^0-\alpha )=0$

• A. Both A and R are true and R is the correct explanation to A.
• B. Both A & R are true but R is not the correct explanation to A.
• C. A is true, R is false.
• D. A is false, R is true.

Answer 1

The correct option is A Both A and R are true and R is the correct explanation to A.

Assertion (A) :
${\sin }^2\theta +{\sin }^2(\theta +60)+{\sin }^2(\theta -60)={\sin }^2\theta +\left(\sin \right(\theta +60)+\sin (\theta -60){)}^2-2\sin (\theta +60\left)\sin \right(\theta -60)$
$={\sin }^2\theta +(2\sin \theta \cos 60{)}^2-[-\cos 2\theta +\cos \left(120\right)]$
$=2{\sin }^2\theta -[-2{\cos }^2\theta +1-\frac{1}{2}]=2{\sin }^2\theta +\cos 2\theta +\frac{1}{2}$
$=\frac{3}{2}$

Reason (R):
$\cos \alpha +\cos (120+\alpha )+\cos (120-\alpha )$
$\Rightarrow \cos \alpha +2\cos \left(120\right)\cos \alpha [\cos C+\cos D=2\cos (\frac{C+D}{2}\left)\cos \right(\frac{C-D}{2}\left)\right]$
$\Rightarrow \cos \alpha -\cos \alpha$
$=0$
Both A and R are True and R is a correct explanation to A.