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