Trigonometry
Trigonometric Identities
GRB_1000_SCQ
Grade Class 11

Question:

If $p = \cos 55°$, $q = \cos 65°$ and $r = \cos 175°$, then the value of $\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{r}{pq}$ is equal to:
0
-1
1
2

Step-by-Step Solution

Key Concept: Sum-to-product formulas for cosines and properties of cosine function.
Step 1: Rewrite the expression with a common denominator. We need to find $\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{r}{pq}$. To combine these fractions, we express them with the common denominator $pq$: $$\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{r}{pq} = \dfrac{q}{pq} + \dfrac{p}{pq} + \dfrac{r}{pq} = \dfrac{p + q + r}{pq}$$ Step 2: Substitute the given values. Now we substitute $p = \cos 55°$, $q = \cos 65°$, and $r = \cos 175°$: $$\dfrac{p + q + r}{pq} = \dfrac{\cos 55° + \cos 65° + \cos 175°}{pq}$$ Step 3: Simplify $\cos 55° + \cos 65°$ using the sum-to-product formula. We use the formula $\cos A + \cos B = 2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)$: $$\cos 55° + \cos 65° = 2\cos\left(\dfrac{55° + 65°}{2}\right)\cos\left(\dfrac{65° - 55°}{2}\right)$$ $$= 2\cos(60°)\cos(5°) = 2 \cdot \dfrac{1}{2} \cdot \cos 5° = \cos 5°$$ Step 4: Simplify $\cos 175°$ using the supplementary angle identity. We recognize that $175° = 180° - 5°$, so: $$\cos 175° = \cos(180° - 5°) = -\cos 5°$$ Step 5: Find the sum $p + q + r$. Combining the results from Steps 3 and 4: $$p + q + r = \cos 55° + \cos 65° + \cos 175° = \cos 5° + (-\cos 5°) = 0$$ Step 6: Calculate the final answer. Since the numerator equals zero: $$\dfrac{p + q + r}{pq} = \dfrac{0}{pq} = 0$$ Therefore, the value of $\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{r}{pq}$ is $\boxed{0}$. The answer is **Option 1: 0**.
Correct Answer: 1

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