Question

If $p=\cos {55}^{\circ }$ , $q=\cos {65}^{\circ }$ and $r=\cos {175}^{\circ }$ then the value of $\frac{1}{p}+\frac{1}{q}+\frac{r}{pq}$ is

• A. $0$
• B. $-1$
• C. $1$
• D. None of these

Answer 1

The correct option is A

$0$

Step-1: Apply L.C.M.

Given $p=\cos {55}^{\circ }$, $q=\cos {65}^{\circ }$and $r=\cos {175}^{\circ }$

To find the value of $\frac{1}{p}+\frac{1}{q}+\frac{r}{pq}$,

Take L.C.M

$\frac{1}{p}+\frac{1}{q}+\frac{r}{pq}$=$\frac{q+p+r}{pq}$

Step-2:Apply Trigonometric Formula:

Consider $\frac{q+p+r}{pq}$

Substitute the values of $p,q,\mathrm{and}r$ in the above term

$\frac{q+p+r}{pq}$= $\frac{\cos {65}^{\circ }+\cos {55}^{\circ }+\cos {175}^{\circ }}{\cos {55}^{\circ }\cos {65}^{\circ }}$

Apply $\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$ formula to the value $\cos {65}^{\circ }+\cos {55}^{\circ }$

and write $\cos {175}^{\circ }=\cos ({180}^{\circ }-5^{\circ })$ in the above equation

$\frac{q+p+r}{pq}$= $\frac{2\cos \left({\displaystyle \frac{{65}^{\circ }+{55}^{\circ }}{2}}\right)+\cos \left(\frac{{65}^{\circ }-{55}^{\circ }}{2}\right)+\cos ({180}^{\circ }-5^{\circ })}{\cos {55}^{\circ }\cos {65}^{\circ }}$

= $\frac{2\cos {60}^{\circ }+\cos 5^{\circ }-\cos 5^{\circ }}{\cos {55}^{\circ }\cos {65}^{\circ }}$ [Since, $\cos ({180}^{\circ }-5^{\circ })=-\cos 5^{\circ }$]

= $\frac{2\left({\displaystyle \frac{1}{2}}\right)+\cos 5^{\circ }-\cos 5^{\circ }}{\cos {55}^{\circ }\cos {65}^{\circ }}$ [Since, $\cos {60}^{\circ }=\frac{1}{2}$]

= $\frac{\cos 5^{\circ }-\cos 5^{\circ }}{\cos {55}^{\circ }\cos {65}^{\circ }}$

$\therefore \frac{q+p+r}{pq}=0$

Hence, the value of $\frac{1}{p}+\frac{1}{q}+\frac{r}{pq}$ is $0$.