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`

Answer 1

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`