If `x=secphi -tanphi and y="cosec" phi+cotphi`, then show that `xy+x-y+1=0.`

Explanation

Given that, `" "x=secphi-tanphi" "...(i)`
and `" "y="cosec"phi+cotphi" "...(ii)`
`" "1*xy=(secphi-tanphi)("cosec"phi+cotphi )`
`rArr" "xy=secphi*"cosec"phi-"cosec"phi*tanphi+secphi*cot phi -tanphi*cotphi`
`rArr" "xy=secphi*"cosec"phi-(1)/(cosphi)+(1)/(sinphi)-1`
`rArr" "1+xy=secphi"cosec"phi-sec phi+"cosec"phi" "...(iii)`
From Eqs. (i) and (ii), we get
`" "x-y=secphi-tanphi-"cosec"phi-cotphi`
`rArr" "x-y=secphi-"cosec"phi-(sinphi)/(cosphi)-(cosphi)/(sinphi)`
`rArr" "x-y=secphi-"cosec"phi-((sin^(2)phi+cos^(2)phi)/(sinphi*cosphi))`
`rArr" "x-y=sec phi-"cosec"phi-"cosec"phi*secphi`
`rArr" " x-y=-(secphi*"cosec"phi-secphi+"cosec"phi)`
`rArr" "x-y=-(xy+1)" "` [from Eq. (iii)]
`rArr" "xy+x-y+1=0" "` Hence proved.