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प्रश्न
Prove that `(sin^2θ)/(cos θ) + cos θ = sec θ`.
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उत्तर
L.H.S. = `(sin^2θ)/(cos θ) + cos θ`
= `(sin^2θ + cos^2θ)/(cos θ)`
= `1/(cos θ)` ...[∵ sin2θ + cos2θ = 1]
= sec θ
= R.H.S.
∴ `(sin^2θ)/(cos θ) + cos θ = sec θ`
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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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and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
