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If 3√2sin θ + √2 = √2cos2 θ + 3 + sinθ , 0° < θ < 90°, then the value of (tan2 θ + cosec2 θ – sec2 θ + 2) is∶
1. 1
2. 2
3. 3
4. 4

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Best answer
Correct Answer - Option 3 : 3

Given :

3√2sin θ + √2 = √2cos2 θ + 3 + sinθ

Formula used :

sin2θ + cos2θ = 1

Calculations : 

3√2sin θ + √2 = √2cos2 θ + 3 + sinθ

⇒ 3√2sin θ + √2 = √2(1 - sin2 θ) + 3 + sinθ

⇒ 3√2sin θ + √2 = √2 - √2sin2 θ + 3 + sinθ

⇒ 3√2sin θ + √2sin2 θ  = 3 + sinθ

⇒ √2sin2 θ + 3√2sin θ  = 3 + sinθ

⇒ √2sin2 θ + 3√2sin θ - sin θ - 3 = 0

⇒ √2sin θ(sin θ + 3) - 1(sin + 3) = 0

⇒ sin θ = 1/√2 or -3 

⇒ θ = 45° 

Now 

(tan2 θ + cosec2 θ – sec2 θ + 2) = 12 + √22 - √22 + 2

⇒ 1 + 2 - 2 + 2 

⇒ 3 

∴ The correct choice will be option 3.

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