Question

I: $4(\sin {24}^{\circ }+\cos 6^{\circ })=\sqrt{15}+\sqrt{3}$

II: $\sin {21}^{\circ }\cos 9^{\circ }-\cos {84}^{\circ }\cos 6^{\circ }=\frac{1}{4}$

• A. only I is true
• B. only II is true
• C. Both I and II are true
• D. Neither I nor II are true

Answer 1

Answer: (C)

Both I and II are true

I: $4(\sin 24+\cos 6)=4(\sin 24+\cos (90-84))$

$=4(\sin 24+\sin 84)=8\sin 54\cos 30$

$\frac{8\sqrt{3}}{2}\left(\frac{\sqrt{5}+1}{4}\right)=\sqrt{15}+\sqrt{3}$

II: $\sin 21\cos 9-\cos 84\cos 6$

$=\frac{1}{2}(\sin 30+\sin 12-\cos 90-\cos 78)$

$=\frac{1}{2}(\sin 30+\sin 12-\cos 90-\cos (90-12))$

$=\frac{1}{2}(\sin 30+\sin 12-\cos 90-\sin 12)$

$=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

Hence, option ${}^{\prime }{C}^{\prime }$ is correct.