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If $x = sec \phi - \tan \phi \: \: and \: \: y = cosec \phi + cot \phi$ , then show that xy + x – y + 1 = 0. [Hint: Find $xy + 1$ and then show tan $x - y = -(xy + 1)$ ]

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$\\ x=sec \phi - \tan \phi ~~~y=cosec \phi +cot \phi \\\\ L.H.S=xy+x - y+1 \\\\ = \left( sec \phi - \tan \phi \right) \left( cosec \phi +cot \phi \right) + \left( sec \phi - \tan \phi \right) - \left( cosec \phi +cot \phi \right) +1 \\\\ = \left( \frac{1}{cosec \phi } - \frac{\sin \phi }{\cos \phi } \right) \left( \frac{1}{\sin \phi }+\frac{\cos \phi }{\sin \phi } \right) + \left( \frac{1}{\cos \phi } - \frac{\sin \phi }{\cos \phi } \right) - \left( \frac{1}{\sin \phi }+\frac{\cos \phi }{\sin \phi } \right) \\+1 \\\\$

$\\ = \left( \frac{1 - \sin \phi }{\cos \phi } \right) \left( \frac{1+\cos \phi }{\sin \phi } \right) +\frac{1 - \sin \phi }{\cos \phi } - \frac{1+\cos \phi }{\sin \phi }+1~~$

$=\frac{1 - \sin \phi +\cos \phi - \sin \phi \cos \phi }{\cos \phi \sin \phi }+\frac{\sin \phi - \sin ^{2} \phi - \cos \phi - \cos ^{2} \phi }{\cos \phi \sin \phi} +1 \\\\$

$\\ =\frac{1 - \sin \phi +\cos \phi - \sin \phi \cos \phi +\sin \phi - \sin ^{2} \phi - \cos \phi - \cos ^{2} \phi +\cos \phi \sin \phi }{\cos \phi \sin \phi } \\\\ =\frac{1 - 1}{\cos \phi \sin \phi }=0 \\\\ L.H.S= R.H.S \\\\$

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