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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. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
