Q61 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics
$\begin{aligned} & \left(\cos 252^{\circ}-\sin 126^{\circ}\right)\left(\cos 252^{\circ}+\right. \\ & \left.\sin 126^{\circ}\right)\left(\sin ^2 126^{\circ}+\sin ^2 186^{\circ}+\sin ^2 66^{\circ}\right)=\end{aligned}$
- A$\frac{3 \sqrt{5}}{8}$
- Correct answerB$\frac{-3 \sqrt{5}}{8}$
- C$\frac{-3 \sqrt{5}}{4}$
- D$\frac{3 \sqrt{5}}{4}$
Solution
$\begin{aligned} & \left(\cos 252^{\circ}-\sin 126^{\circ}\right)\left(\cos 252^{\circ}+\sin 126^{\circ}\right) \\ & \left(\sin ^2 126^{\circ}+\sin ^2 186^{\circ}+\sin ^2 66^{\circ}\right) \\ & =\left(\cos ^2 252^{\circ}-\sin ^2 126^{\circ}\right)\left(\sin ^2 126^{\circ}+\sin ^2 186^{\circ}+\sin ^2 66^{\circ}\right) \\ & =\left[\frac{1+\cos 504^{\circ}}{2}-\frac{1-\cos 252^{\circ}}{2}\right] \\ & {\left[\frac{1-\cos 252^{\circ}+1-\cos 372^{\circ}+1-\cos 132^{\circ}}{2}\right]} \\ & =\frac{1}{4}\left(\cos 504^{\circ}+\cos 252^{\circ}\right) \\ & {\left[3-\cos 252^{\circ}-2 \cos 252^{\circ} \cos 120^{\circ}\right]} \\ & =\frac{1}{4}\left(2 \cos 378^{\circ} \cos 126^{\circ}\right)\left[3-\cos 252^{\circ}+\cos 252^{\circ}\right] \\ & =\frac{3}{2} \cos 18^{\circ}\left(-\sin 36^{\circ}\right)=-\frac{3}{2} \times \frac{\sqrt{10+2 \sqrt{5}}}{4} \times \frac{\sqrt{10-2 \sqrt{5}}}{4} \\ & =-\frac{3}{32} \sqrt{100-20}=\frac{-3}{32} \sqrt{80}=-\frac{3}{32}(4 \sqrt{5}) \\ & =-\frac{3 \sqrt{5}}{8} \\ & \end{aligned}$
Hence, the correct answer is option: 'B'.