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 tan x – y = –(xy + 1)]
According to the question,
x = sec ϕ – tan ϕ and y = cosec ϕ + cot ϕ
Given that, LHS = xy + x – y + 1
$=(\sec \phi-\tan \phi)(\operatorname{cosec} \phi+\cot \phi)+(\sec \phi-\tan \phi)-(\operatorname{cosec} \phi+\cot \phi)+1$
$=\sec \phi \operatorname{cosec} \phi+\cot \phi \sec \phi-\tan \phi \cot \phi-\tan \phi \operatorname{cosec} \phi$ $+\sec \phi-\tan \phi-(\operatorname{cosec} \phi+\cot \phi)+1$
$=\frac{1}{\sin \phi \cos \phi}+\frac{1}{\sin \phi}-1-\sec \phi+\sec \phi-\tan \phi-\left(\frac{1}{\sin \phi}+\frac{\cos \phi}{\sin \phi}\right)+1$
$=\frac{1}{\sin \phi \cos \phi}+\frac{1}{\sin \phi}-\tan \phi-\left(\frac{1}{\sin \phi}+\frac{\cos \phi}{\sin \phi}\right)$
$=\frac{1}{\sin \phi \cos \phi}-\frac{\cos \phi}{\sin \phi}-\frac{\sin \phi}{\cos \phi}$
$=\frac{1}{\sin \phi \cos \phi}-\left(\frac{\cos \phi}{\sin \phi}+\frac{\sin \phi}{\cos \phi}\right)$
$=\frac{1}{\sin \phi \cos \phi}-\left(\frac{\cos ^{2} \phi+\sin ^{2} \phi}{\sin \phi \cos \phi}\right)$
Since, $\sin ^{2} \theta+\cos ^{2} \theta=1$
$=\frac{1}{\sin \phi \cos \phi}-\left(\frac{1}{\sin \phi \cos \phi}\right)=0$
Thus, LHS = xy + x – y + 1 = 0