if tan2 α = 3 + Q2, the sec α + tan3 α cosec α = ?

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SSC CGL 2022 Tier-I Official Paper (Held On : 07 Dec 2022 Shift 1)
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  1. \((3+Q^2)^\frac{3}{2}\)
  2. \((7+Q^2)^\frac{3}{2}\)
  3. \((5-Q^2)^\frac{3}{2}\)
  4. \((4+Q^2)^\frac{3}{2}\)

Answer (Detailed Solution Below)

Option 4 : \((4+Q^2)^\frac{3}{2}\)
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Detailed Solution

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Given:

tan2 \(α\) = 3 + Q2

Formula used:

Sec2 \(α\) - tan2 \(α\) = 1

Calculations:

⇒ tan2 \(α\) = 3 + Q2

⇒ Sec2 \(α\) - 1  = 3 + Q2

⇒ Sec2 \(α\) = 4 + Q2

⇒ Sec \(α\) = \(√{4+Q^2}\)

The given equation is 

⇒ Sec \(α\)  + tan3 \(α\) cosec \(α\)

\(\frac{1}{cos α} + \frac{ sin^3 α}{cos^3 α}×\frac{1}{sin α}\)

\(\frac{1}{cos α} + \frac{ sin^2 α}{cos^3 α} \)

\(\frac{cos^2α +sin^2 α}{cos^3 α} \)

\(\frac{1}{cos^3 α} \)

⇒ Sec3 \(α\)

(Sec \(α\))3

\((√{4+Q^2})^3\)

\((4+Q^2 )^{\frac{3}{2}}\)

⇒ Hence, Option 4 is correct

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