For the following two (02) items: Let p = cosec\, 20° and q = cosec\, 70° . What is (p^2 + q^2)/(p^2 q^2) equal to?

Correct answer: B. 1 Given p = cosec\, 20^ ^ and q = cosec\, 70^ ^ . The given expression is (p^2 + q^2)/(p^2 q^2) . This can be rewritten as (1)/(q^2) + (1)/(p^2) . Substituting the values of p and q : 1 cosec^2 70^ ^ + 1 cosec^2 20^ ^ = ^2 70^ ^ + ^2 20^ ^ Since 70^ ^ = (90^ ^ - 20^ ^ ) = 20^ ^ , we get: = ^2 20^ ^ + ^2 20^ ^ = 1 Answer: 1

NDA 2025

Exam: NDA
Year: 2025
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NDA 2025 · Mathematics · Trigonometric Ratios & Identities

For the following two (02) items: Let p = cosec\, 20° and q =...

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?

Options

A $\dfrac{1}{2}$
B $1$
C $\dfrac{3}{2}$
D $2$

Correct answer

B. $1$

Step-by-step 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

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