If P cos α = 3 and 4 tan α = Q, then what is the relation between P and Q, which is independent of α?  

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SSC CGL 2023 Tier-I Official Paper (Held On: 19 Jul 2023 Shift 2)
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  1. \(\rm \frac{9}{p^2}+\frac{16}{Q^2}=1\)
  2. \(\rm \frac{9}{p^2}-\frac{16}{Q^2}=1\)
  3. \(\rm \frac{p^2}{9}-\frac{Q^2}{16}=1\)
  4. \(\rm \frac{p^2}{9}+\frac{Q^2}{16}=1\)

Answer (Detailed Solution Below)

Option 3 : \(\rm \frac{p^2}{9}-\frac{Q^2}{16}=1\)
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Detailed Solution

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Concept used:

sec2 θ - tan2 θ = 1

Calculation:

P cos α = 3

⇒ cos α = 3/P

⇒ sec α = P/3

4 tan α = Q

⇒ tan α = Q/4

From the given concept, we get

sec2 α - tan2 α = 1

⇒ P2/9 - Q2/16 = 1

∴ The correct answer is option 3

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