cosecA = 1/sinA, so sinA = 2/7.

Also, secA = 1/cosA. cosA = +- sqrt(1 - (sinA)^2) = +- sqrt(1 - (2/7)^2) = +- sqrt(45/49) =

= +- (3/7)*sqrt(5).

Now we have to choose sign for cosA. For 90<A<180 cosA < 0, so "-".

So secA = 1/cosA = -[7/(3*sqrt(5))] = **-(7/15)*sqrt(5)**.

90<A<180 so cosA<0

cosecA=` 7/2 =>sinA=2/7`

`cosA=+-sqrt(1-(sin^2A))`

` cosA=-(3sqrt(5))/7 `

`:.secA=1/cosA=-7/(3sqrt(5))`

``

` `

This is simple and is explained with procedural steps it is as follows

cosec A = 7/2, where 90<A<180 ie `Pi/2< A <Pi`

`=>`

A = `pi -sin^(-1)(2/7)`

`A= pi - cos^(-1)((3sqrt(5))/7) `

`now`

taking sec on both sides

ie

`sec A=sec( pi - cos^(-1)((3sqrt(5))/7))`

= `1/cos( pi - cos^(-1)((3sqrt(5))/7))`

we know ` cos(Pi)=-1`

`=>`

in the above equation the denominator is of the form

cos(a-b)=cos a cos b + sin a sin b

where a= pi and b= `cos^(-1)((3sqrt(5))/7)`

`so,`

` sec A=1/cos( pi - cos^(-1)((3sqrt(5))/7))`

`=1/((cos(pi)*cos(cos^(-1)((3sqrt(5))/7))+sin(pi)*(sin^(-1)((3sqrt(5))/7) ))`

`= 1/(-1*(3sqrt(5))/7)+0 `

`=(-7)/(3sqrt(5))`