Question

Match the following.
List - I List - II
1. $\sin {48}^{\circ }\sin {12}^{\circ }$ a.$\frac{\sqrt{3}}{\sqrt{2}}$
2. $\cos {48}^{\circ }\cos {12}^{\circ }$ b.$\frac{\sqrt{15}+\sqrt{3}}{4}$
3. $\sin {75}^{\circ }+\sin {15}^{\circ }$ c.$\frac{\sqrt{5}-1}{8}$
4 $\sin {24}^{\circ }+\cos 6^{\circ }$ d.$\frac{3+\sqrt{5}}{8}$

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

Answer 1

Answer: (C)

$1-c,2-d,3-a,4-b$

$1.\sin {48}^{\circ }\sin {12}^{\circ }=\frac{\cos \left({36}^{\circ }\right)-\cos \left({60}^{\circ }\right)}{2}=\frac{\frac{\sqrt{5}+1}{4}-\frac{1}{2}}{2}=\frac{\sqrt{5}-1}{8}$
$[\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}and\sin {18}^{\circ }=\frac{\sqrt{5}-1}{4}]$
$2.\cos {48}^{\circ }\cos {12}^{\circ }=\frac{\left[\cos \right({60}^{\circ }+\cos \left({36}^{\circ }\right)\left)\right]}{2}=\frac{\frac{1}{2}+\frac{\sqrt{5}+1}{4}}{2}=\frac{\sqrt{5}+3}{8}$
$3.\sin {75}^{\circ }+\sin {15}^{\circ }=2\sin \left(\frac{{90}^{\circ }}{2}\right)\cos \left({30}^{\circ }\right)=2\times \frac{1}{\sqrt{2}\times \frac{\sqrt{3}}{2}}=\sqrt{\frac{3}{2}}$
$4.\sin {24}^{\circ }+\cos 6^{\circ }=(\sin {24}^{\circ }+\sin {84}^{\circ })=2\sin \left({54}^{\circ }\right)\cos {30}^{\circ }=\sqrt{3}\left[\sin {54}^{\circ }\right]$
$=\sqrt{3}\cos {36}^{\circ }=\sqrt{3}\frac{(\sqrt{5}+1)}{4}$
$=\frac{\sqrt{15}+\sqrt{3}}{4}$
Hence, option 'C' is correct.