If `sin theta+ cos theta = p` and `sec theta + cosec theta = q`; show that `q(p^2-1) = 2p`

Explanation

Given, that, `sintheta+costheta=p` ……………(i)
and `sectheta-cosectheta=q`
`rArr 1/(costheta) + 1/(sintheta) =q` `[therefore sectheta=1/(costheta)` and `"cosec"theta=1/(sintheta)]`
`rArr (sintheta + costheta)/(sintheta.costheta)=q`
`rArr (p)/(sintheta.costheta)=q`
`rArr sintheta. costheta = p/q` [From Eq. (i)......(ii)]
`sintheta + costheta=p`
On squaring both the sides, we get
`(sintheta + costheta)^(2)=p^(2)`
`rArr sin^(2)theta+cos^(2)theta+2sintheta. costheta=p^(2)` `[therefore(a+b)^(2)=a^(2)+2ab+b^(2)]`
`rArr 1+2sintheta. costheta=p^(2)` (From Eq. (ii))
`rArr q+2p=p^(2)q rArr 2p=p^(2)q-q`
`rArr q(p^(2)-1)=2p` Hence Proved.