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Question
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
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Solution
L.H.S = `(sin theta)/(1 - cot theta) + (cos theta)/(1- tan theta)`
`= (sin theta)/(1 - cos theta/sin theta) + cos theta/(1 - sin theta/cos theta)`
`= sin^2 theta/(sin theta - cos theta) + cos^2 theta/(cos theta - sin theta)`
`= (sin^2 theta)/(sin theta - cos theta) - cos^2 theta/(sin theta - costheta)`
`= (sin^2 theta - cos^2 theta)/(sin theta - cos theta)`
`= ((sin theta - cos theta)(sin theta + cos theta))/(sin theta - cos theta)`
`= sin theta + cos theta`
= R.H.S
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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`
