If `secalpha-tanalpha=p` and `cosec alpha+cotalpha=q`, then express p in terms of q and also q in terms p.

Explanation

We have ,
`secalpha=p+tanalpharArrsec^(2)alpha=(p+tanalpha)^(2)`
`rArr1+tan^(2)alpha=p^(2)+tan^(2)alpha+2ptanalpha`
`rArrtanalpha=(1-p^(2))/(2p)` …(1)
Also , `"cosec" alpha=q-cotalpharArr"cosec"^(2)alpha=(q-cotalpha)^(2)`
`1+cot^(2)alpha=q^(2)+cot^(2)alpha=2pcotalpha`
`cotalpha=(q^(2)-1)/(2p)rArrtanalpha=(2q)/(q^(2)-1)` ...(2)
From equations (1) and (2), we get
`(1-p^(2))/(2p)=(2p)/(q^(2)-1)`
`rArrq^(2)-1-p^(2)q^(2)+p^(2)=4pq`
`rArrp^(2)q^(2)+1+2pq=q^(2)+p^(2)-2pq`
`rArr(pq+1)^(2)=(q-p)^(2)`
`rArrpq+1=+-(q-p)`
`rArrpq+1=q-p`
`rArrp(q+1)=q-1`
`rArrp=(q-1)/(q+1)`
and `q(p-1)=-p-1`
`rArrq=(1+p)/(1-p)`
`rArrpq+1=-q+p`
`rArrp(q-1)=q-1`
`rArrp=(1+q)/(1-q)`
and `q(p+1)=p-1`
`:.q=(p-1)/(p+1)`
So , if `p=(q-1)/(q+1)`,then`q=(1+p)/(1-p)`
and if `p=(1+q)/(1-q)` ,then `q=(p-1)/(p+1)`