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प्रश्न
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
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उत्तर
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2`
= `(sinθ/cosθ + 1/cosθ)^2 + (sinθ/cosθ - 1/cosθ)^2`
= `((sinθ + 1)/cosθ)^2 + ((sinθ - 1)/cosθ)^2`
= `(sinθ + 1)^2/(cos^2θ) + (sinθ - 1)^2/cos^2θ`
= `((sinθ + 1)^2 + (sinθ - 1)^2)/cos^2A`
= `(sin^2θ + 1 + 2sinθ + sin^2θ + 1 - 2sinθ)/(1 - sin^2θ)`
= `(2(1 + sin^2θ))/(1 - sin^2θ)`
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We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
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∴ sin θ = `9/41`
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