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प्रश्न
Prove that cot2θ × sec2θ = cot2θ + 1
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उत्तर
L.H.S = cot2θ × sec2θ
= `(cos^2theta)/(sin^2theta) xx 1/(cos^2theta)`
= `1/(sin^2theta)`
= cosec2θ
= 1 + cot2θ ......[∵ 1 + cot2θ = cosec2θ]
= R.H.S
∴ cot2θ × sec2θ = cot2θ + 1
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`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
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We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
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`square/square` = cosec2θ ......[Taking root on the both side]
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∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
