(sin^4x-cos^4x+1)cosec^2x=2
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❤️❤️(sin⁴θ-cos⁴θ+1)cosec²θ
=[{(sin²θ)²-(cos²θ)²}+1]cosec²θ
=[{(sin²θ+cos²θ)(sin²θ-cos²θ)}+1]cosec²θ
=(sin²θ-cos²θ+1)cosec²θ [∵, sin²θ+cos²θ=1]
={sin²θ+(1-cos²θ)}cosec²θ
=(sin²θ+sin²θ)cosec²θ
=2sin²θ.cosec²θ
=2sin²θ×1/sin²θ
=2 (Proved)❤️❤️
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