Question

Match the following.
List - I List - II
1.$\cos 6^{\circ }\sin {24}^{\circ }\cos {72}^{\circ }$ a.$\frac{3}{16}$

2. $\cos {10}^{\circ }\cos {30}^{\circ }\cos {50}^{\circ }\cos {70}^{\circ }$ b.$\frac{1}{16}$

3.$\sin 2^{\circ }\sin {24}^{\circ }\sin {48}^{\circ }\sin {84}^{\circ }$ c.$\frac{3}{16}$

4. $\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }$ d.$\frac{1}{8}$

• A. $1-c,2-b,3-a,4-d$
• B. $1-d,2-c,3-b,4-a$
• C. $1-d,2-b,3-a,4-c$
• D. $1-b,2-c,3-d,4-a$

Answer 1

The correct option is B $1-d,2-c,3-b,4-a$

$A)$ $\cos 6^{\circ }\sin {24}^{\circ }\cos {72}^{\circ }=\cos 6^{\circ }\cos {66}^{\circ }\cos {72}^{\circ }$

$=\frac{\cos 6^{\circ }\cos {66}^{\circ }\cos {54}^{\circ }\cos {72}^{\circ }}{\cos {54}^{\circ }}$

$(\because \cos \theta \cos ({60}^{\circ }-\theta \left)\cos \right({60}^{\circ }+\theta )=\frac{1}{4}\cos 3\theta )$

$=\frac{\cos (3\times 6^{\circ })\cos {72}^{\circ }}{8\cos {54}^{\circ }}=\frac{cos{18}^{\circ }\cos {72}^{\circ }}{cos{54}^{\circ }}$

$=\frac{\cos ({18}^{\circ }+{72}^{\circ })+\cos ({18}^{\circ }-{72}^{\circ })}{8\cos {54}^{\circ }}=\frac{\cos {90}^{\circ }+\cos {54}^{\circ }}{\cos {54}^{\circ }}=\frac{1}{8}$

$B)$ $\cos {10}^{\circ }\cos {30}^{\circ }\cos {50}^{\circ }\cos {70}^{\circ }=\cos {30}^{\circ }\cos {10}^{\circ }\cos {50}^{\circ }\cos {70}^{\circ }$

$(\because \cos \theta \cos ({60}^{\circ }-\theta \left)\cos \right({60}^{\circ }+\theta )=\frac{1}{4}\cos 3\theta )$

$=\frac{\sqrt{3}}{2\times 4}\cos (3\times {10}^{\circ })=\frac{\sqrt{3}}{8}\cos {30}^{\circ }=\frac{3}{16}$

$C)$ $\sin {12}^{\circ }\sin {24}^{\circ }\sin {48}^{\circ }\sin {84}^{\circ }$

$=\frac{1}{4}\left(2\sin {12}^{\circ }\sin {48}^{\circ }\right)\left(2\sin {24}^{\circ }\sin {84}^{\circ }\right)$

$=\frac{1}{4}\left(\cos \right({12}^{\circ }-{48}^{\circ })-\cos ({12}^{\circ }+{48}^{\circ }\left)\right)\left(\cos \right({24}^{\circ }-{84}^{\circ })-\cos ({24}^{\circ }+{84}^{\circ }\left)\right)$

$=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

$D)$$\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }$

$(\because \sin \theta \sin ({60}^{\circ }-\theta \left)\sin \right({60}^{\circ }+\theta )=\frac{1}{4}\sin 3\theta )$

$=\frac{1}{4}\sin {60}^{\circ }\sin (3\times {20}^{\circ })=\frac{1}{4}(\sin {60}^{\circ }{)}^2=\frac{3}{16}$