Q178 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following two (02) items:
Let $p = \text{cosec}\, 20°$ and $q = \text{cosec}\, 70°$.
What is $\dfrac{p^2 + q^2}{p^2 q^2}$ equal to?
Let $p = \text{cosec}\, 20°$ and $q = \text{cosec}\, 70°$.
What is $\dfrac{p^2 + q^2}{p^2 q^2}$ equal to?
- A$\dfrac{1}{2}$
- Correct answerB$1$
- C$\dfrac{3}{2}$
- D$2$
Solution
Given $p = \text{cosec}\, 20^\circ$ and $q = \text{cosec}\, 70^\circ$.
The given expression is $\dfrac{p^2 + q^2}{p^2 q^2}$.
This can be rewritten as $\dfrac{1}{q^2} + \dfrac{1}{p^2}$.
Substituting the values of $p$ and $q$:
$\dfrac{1}{\text{cosec}^2 70^\circ} + \dfrac{1}{\text{cosec}^2 20^\circ}$
$= \sin^2 70^\circ + \sin^2 20^\circ$
Since $\sin 70^\circ = \sin(90^\circ - 20^\circ) = \cos 20^\circ$, we get:
$= \cos^2 20^\circ + \sin^2 20^\circ$
$= 1$
Answer: $1$
The given expression is $\dfrac{p^2 + q^2}{p^2 q^2}$.
This can be rewritten as $\dfrac{1}{q^2} + \dfrac{1}{p^2}$.
Substituting the values of $p$ and $q$:
$\dfrac{1}{\text{cosec}^2 70^\circ} + \dfrac{1}{\text{cosec}^2 20^\circ}$
$= \sin^2 70^\circ + \sin^2 20^\circ$
Since $\sin 70^\circ = \sin(90^\circ - 20^\circ) = \cos 20^\circ$, we get:
$= \cos^2 20^\circ + \sin^2 20^\circ$
$= 1$
Answer: $1$
Hence, the correct answer is option: 'B'.