Question

Solve : $\frac{\cos {10}^o}{\sin {80}^o}+\sin {42}^o\times 4\sec {48}^o$

• A. $3$
• B. $2$
• C. $5$
• D. $4$

Answer 1

The correct option is A $3$

$\frac{\cos {10}^{\circ }}{\sin {80}^{\circ }}+\sin {42}^{\circ }\times 4\sec {48}^{\circ }$

$\frac{\cos {10}^{\circ }}{\sin (90-10{)}^{\circ }}+\sin {42}^{\circ }\times 4\sec (90-42{)}^{\circ }$

$=\frac{\cos {10}^{\circ }}{\cos {10}^{\circ }}+\sin {42}^{\circ }\times 4\csc {42}^{\circ }\dots \left(\sin \right(90-x)=cosx)and\left(\sec \right(90-x)=cscx)$

$=\frac{\cos {10}^{\circ }}{\cos {10}^{\circ }}+4(\sin {42}^{\circ }\times \csc {42}^{\circ })$

$=1+4\left(1\right)$

$\therefore \frac{\cos {10}^{\circ }}{\sin {80}^{\circ }}+\sin {42}^{\circ }\times 4\sec {48}^{\circ }$$=5$