Question

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

Answer 1

Hint: Here, we will first take the LCM of the given expression. Then we will substitute the given values in the expression. We will then simplify the expression using trigonometric formulas and identities. We will solve the equation further to get the required value.

Formula Used:
 $\cos A+\cos B=2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)$

Complete step-by-step answer:
It is given that $p=\cos {55}^{\circ }$, $q=\cos {65}^{\circ }$ and $r=\cos {175}^{\circ }$.
First, we will take the LCM of the denominators of the given expression ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{r}{pq}}$, we get
${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{r}{pq}}={\displaystyle \frac{q+p+r}{pq}}$
Now, substituting the given values in this fraction, we get,
$\Rightarrow {\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{r}{pq}}={\displaystyle \frac{\cos {65}^{\circ }+\cos {55}^{\circ }+\cos {175}^{\circ }}{\cos {55}^{\circ }\times \cos {65}^{\circ }}}$
Now, using the formula $\cos A+\cos B=2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)$ in the numerator for $\cos {65}^{\circ }+\cos {55}^{\circ }$, we get,
$\Rightarrow {\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{r}{pq}}={\displaystyle \frac{2\cos {60}^{\circ }\cos 5^{\circ }+\cos \left({180}^{\circ }-5^{\circ }\right)}{\cos {55}^{\circ }\times \cos {65}^{\circ }}}$
Now, as we know, $\cos \left({180}^{\circ }-\theta \right)=-\cos \theta$ because in the second quadrant, cosine is negative. Therefore, we get
$\Rightarrow {\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{r}{pq}}={\displaystyle \frac{2\cos {60}^{\circ }\cos 5^{\circ }-\cos 5^{\circ }}{\cos {55}^{\circ }\times \cos {65}^{\circ }}}$
Now, substituting $\cos {60}^{\circ }={\displaystyle \frac{1}{2}}$ in the above equation, we get
$\Rightarrow {\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{r}{pq}}={\displaystyle \frac{2\times {\displaystyle \frac{1}{2}}\times \cos 5^{\circ }-\cos 5^{\circ }}{\cos {55}^{\circ }\times \cos {65}^{\circ }}}$
$\Rightarrow {\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{r}{pq}}={\displaystyle \frac{\cos 5^{\circ }-\cos 5^{\circ }}{\cos {55}^{\circ }\times \cos {65}^{\circ }}}$
Subtracting the terms in the numerator, we get
${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{r}{pq}}={\displaystyle \frac{0}{\cos {55}^{\circ }\times \cos {65}^{\circ }}}=0$
Therefore, the value of ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{r}{pq}}$ is 0.

Note: Trigonometry is a branch of mathematics that helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine and the cosine function. In simple terms, they are written as ‘sin’, ‘cos’ and ‘tan’. Hence, trigonometry is not just a chapter to study, in fact, it is being used in everyday life.