Question

If $ p=\cos {{55}^{o}},q=\cos {{65}^{o}} $ and $ r=\cos {{175}^{o}}, $ 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

Answer: (A)

Given that
$p=\cos 55^{\circ}, q=\cos 65^{\circ}$ and $r=\cos 175^{\circ}$
Then, $ \frac{1}{p}+\frac{1}{q}+\frac{r}{p q}=\frac{p+q+r}{p q}$
$=\frac{\cos 55^{\circ}+\cos 65^{\circ}+\cos 175^{\circ}}{\cos 55^{\circ} \cos 65^{\circ}}$
$=\frac{\cos 55^{\circ}+2 \cos \left(\frac{175^{\circ}+65^{\circ}}{2}\right) \cos \left(\frac{175^{\circ}-65^{\circ}}{2}\right)}{\cos 55^{\circ} \cos 65^{\circ}}$
$=\frac{\cos 55^{\circ}+2 \cos 120^{\circ} \cos 55^{\circ}}{\cos 55^{\circ} \cos 65^{\circ}}$
$=\frac{\cos 55^{\circ}\left(1+2 \cos 120^{\circ}\right)}{\cos 55^{\circ} \cos 65^{\circ}}$
$=\frac{\left(1-2 \times \frac{1}{2}\right)}{\cos 65^{\circ}} \,\,\left(\because \cos 120^{\circ}=-\frac{1}{2}\right)$
$=\frac{0}{\cos 65^{\circ}}=0$