If `cot theta = (40)/(9)`, find the values of `cosec theta and sin theta`.

Explanation

`1 + cot^2theta = cosec^2theta`
`therefore 1 + ((40)/(9))^2 = cosec^2theta `
` therefore (81+1600)/(81) = cosec^2theta`
`therefore 1 + (1600)/(81) = cosec^2 theta`
`therefore (1681)/(81) = cosec^2theta`
`therefore cosectheta = (41)/(9)`
..... (Taking square roots of both the sides ) ltbgt `sin theta = (1)/(cosec theta)`
`therefore sin theta = (1)/(((41)/(9)))`
`therefore sin theta = (9)/(41)`
`cosec theta = (41)/(9) and sin theta = (9)/(41)`.
Alternative method.
`cot theta = (40)/(9)` ....(Given)......(1)
Consider `trianglePQR` as shown, where `anglePQR = 90^@`
and `anglePQR = theta`.
`cot theta = (QR)/(PQ)` ....(2)
`therefore` from (1) and (2), we get,
`(QR)/(PQ) = (40)/(9)` ......[From (1) and(2)]
`therefore QR = 40k and PQ = 9k `
In `trianglePQR`,
`anglePQR = 90^@` .
`therefore` by Pythagoras theorem,
`PR^2 = PQ^2 + QR^2`
`therefore PR^2 = (9k)^2 + (40k)^2`
`therefore PR^2 = 81k^2 + 1600k^2`
`therefore PR^2 1681k^2`
`therefore` PR = 41k
....(Taking square roots of both the sides)
`cosec theta = (PR)/(PQ) = (41k)/(9k) = (41)/(9)`
`sin theta = (PQ)/(PR)=(9k)/(41k)= (9)/(41)`
` cosec theta = (41)/(9) and sin theta = (9)/(41)`.