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Question
If cot θ = `40/9`, find the values of cosec θ and sinθ,
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`
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Solution
We have, 1 + cot2θ = cosec2θ
1 + `bb((40/9)^2)` = cosec2θ
1 + `bb(1600/81)` = cosec2θ
`(bb81 + bb1600)/bb81` = cosec2θ
`bb1681/bb81` = cosec2θ ......[Taking square root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/bb(41/9)`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
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Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
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= R.H.S.
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