If tan2 θ = 1 - a2, then the value of sec θ + tan3 θ cosec θ is: 

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SSC CGL 2022 Tier-I Official Paper (Held On : 01 Dec 2022 Shift 4)
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  1. \((2-a)^{{3} \over 2}\)
  2. \((a^2 - 1)^{{3} \over 2}\)
  3. \((2 - a^2)^{{3} \over 2}\)
  4. \(a^{{3} \over 2}\)

Answer (Detailed Solution Below)

Option 3 : \((2 - a^2)^{{3} \over 2}\)
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PYST 1: SSC CGL - English (Held On : 11 April 2022 Shift 1)
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Detailed Solution

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Given:

tan2 θ = 1 - a2

Concept Used:

Secθ = 1/cosθ, tanθ = sinθ/cosθ, sinθ = 1/cosecθ

Sec2θ – tan2θ = 1

Calculation:

Secθ = 1/cosθ, tanθ = sinθ/cosθ, sinθ = 1/cosecθ

Sec2θ – tan2θ = 1

We have to calculate, sec θ + tan3 θ cosec θ

\(\begin{array}{l} = \frac{1}{{\cos \theta }} + \frac{{si{n^3}\theta }}{{co{s^3}\theta }} \times \frac{1}{{sin\theta }}\\ = \frac{1}{{\cos \theta }}\left[ {1 + \frac{{si{n^2}\theta }}{{co{s^2}\theta }}} \right]\\ = \frac{1}{{\cos \theta }} \times \frac{1}{{co{s^2}\theta }} = \frac{1}{{co{s^3}\theta }} \end{array}\)

Now, given that,

tan2θ = 1 – a2

⇒ sec2θ – 1 = 1 – a2

\(\begin{array}{l} \Rightarrow \frac{1}{{co{s^2}\theta }} = 2 - {a^2}\\ \Rightarrow \frac{1}{{co{s^3}\theta }} = {\left[ {2 - {a^2}} \right]^{\frac{3}{2}}} \end{array}\)

∴ Option 3 is the correct answer.

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