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प्रश्न
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
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उत्तर
L.H.S. = sec2θ + cosec2θ
= `1/(cos^2θ) + 1/(sin^2θ)`
= `(sin^2θ + cos^2θ)/(cos^2θ.sin^2θ)`
= `1/(cos^2θ.sin^2θ)` ...[∵ sin2θ + cos2θ = 1]
= `1/(cos^2θ) xx 1/(sin^2θ)`
= sec2θ × cosec2θ
= R.H.S.
∴ sec2θ + cosec2θ = sec2θ × cosec2θ
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