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Prove that (sin4 θ − cos4 θ + 1) cosec2 θ = 2.

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L.H.S:

= (sin4 θ - cos4 θ + 1) cosec2 θ

= {(sin2 θ + cos2 θ)(sin2 θ - cos2 θ) + 1} cosec2 θ

= {1(sin2 θ - cos2 θ) + 1} cosec2 θ      [sin2 θ + cos2 θ = 1]

= (sin2 θ - cos2 θ + 1) cosec2 θ

= (sin2 θ - (1 - sin2 θ) + 1) cosec2 θ

= (sin2 θ - 1 + sin2 θ + 1) cosec2 θ

= \(2\, sin^2\, θ\, \times\) \(\frac{1}{sin^2 \theta}\)

= 2

= R.H.S

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