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Question
1 + cot2θ = ?
Options
tan2θ
sec2θ
cosec2θ
cos2θ
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Solution
1 + cot2θ = cosec2θ
Explanation:
`cot^2θ = (cos^2θ)/(sin^2θ)`
So, `1 + cot^2θ = (sin^2θ + cos^2θ)/(sin^2θ)`
= `1/(sin^2θ)`
= cosec2θ
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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]
cosec θ = `41/9`
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`
