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Question
If x = sec Φ – tan Φ and y = cosec Φ + cot Φ then show that xy + x – y + 1 = 0
[Hint: Find xy + 1 and then show that x – y = –(xy + 1)]
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Solution
Given that: x = sec Φ – tan Φ
And y = cosec Φ + cot Φ
xy + x – y + 1 = 0
L.H.S. xy + x – y + 1
= (sec Φ – tan Φ) (cosec Φ + cot Φ) + (sec Φ – tan Φ) – (cosec Φ + cot Φ) + 1
= `(1/cosphi - sinphi/cosphi) (1/sinphi + cosphi/sinphi) + (1/cosphi - sinphi/cosphi) - (1/sinphi - sinphi/cosphi) + 1`
= `((1 - sin phi)/cos phi) ((1 + cos phi)/sinphi) + (1 - sin phi)/cosphi - (1 + cosphi)/sinphi + 1`
= `(1 - sinphi + cosphi - sinphi cosphi)/(cosphisinphi) + (sinphi - sin^2 phi - cos phi - cos^2 phi)/(cos phi sin phi) + 1`
= `(1 - sin phi + cosphi - sinphi cosphi + sinphi - cosphi - (sin^2 phi + cos^2 phi) + sin phi cos phi)/(cosphi sin phi)`
= `(1 - 1)/(cos phi sin phi)`
= 0. R.H.S.
L.H.S. = R.H.S.
Hence proved.
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