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Hint: Try to use fact and \[\sec \theta =\dfrac{1}{\cos \theta }\] and \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and find or evaluate terms of x, y in terms of \[\sin \theta \] and \[\cos \theta \].

__Complete step-by-step answer:__

In the question we are given expressions of x and y in terms of \[\theta \] and we are asked to form an equation or relation between x and y by elimination \[\theta \].

We are given that,

\[x=\sec \theta +\tan \theta \]

We know that \[\sec \theta \] is equal to \[\dfrac{1}{\cos \theta }\] and \[\tan \theta \] is equal to \[\dfrac{\sin \theta }{\cos \theta }\]. So, \[x=\dfrac{1}{\cos \theta }+\dfrac{\sin \theta }{\cos \theta }\] or \[x=\dfrac{1+\sin \theta }{\cos \theta }\]. So, we will multiply the expression with \[\cos \theta \] to numerator and denominator we get,

\[x=\dfrac{\left( 1+\sin \theta \right)\times \cos \theta }{{{\cos }^{2}}\theta }\]

Now, we will use the identity, \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \], so we get x as,

\[x=\dfrac{\left( 1+\sin \theta \right)\cos \theta }{{{\cos }^{2}}\theta }\]

Now, we will write \[\left( 1-{{\sin }^{2}}\theta \right)\] as product of \[\left( 1+\sin \theta \right)\] and \[\left( 1-\sin \theta \right)\]. So, we can write x as,

\[x=\dfrac{\left( 1+\sin \theta \right)\cos \theta }{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}\]

Now on simplification we can say that,

\[x=\dfrac{\cos \theta }{1-\sin \theta }\]

So, x is equal to \[\dfrac{\cos \theta }{1-\sin \theta }\].

Now, we will see expression of y,

\[y=\sec \theta -\tan \theta \]

Now, we write \[\sec \theta \] as \[\dfrac{1}{\cos \theta }\] and \[\tan \theta \] as \[\dfrac{\sin \theta }{\cos \theta }\]. So, we can write y as,

\[y=\dfrac{1}{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }=\dfrac{1-\sin \theta }{\cos \theta }\]

We get, \[y=\dfrac{1-\sin \theta }{\cos \theta }\], which can also be written as, \[y=\dfrac{1}{\dfrac{\cos \theta }{1-\sin \theta }}\].

As we know, \[x=\dfrac{\cos \theta }{1-\sin \theta }\]. So, we can write, \[y=\dfrac{1}{x}\].

Here, on cross multiplication we get, \[xy=1\].

Note: There is another method by using the fact that \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\], which can also be written as, \[\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1\]. So, we know that, \[x=\sec \theta +\tan \theta \] and \[y=\sec \theta -\tan \theta \]. Hence, we can say that, \[xy=1\].

In the question we are given expressions of x and y in terms of \[\theta \] and we are asked to form an equation or relation between x and y by elimination \[\theta \].

We are given that,

\[x=\sec \theta +\tan \theta \]

We know that \[\sec \theta \] is equal to \[\dfrac{1}{\cos \theta }\] and \[\tan \theta \] is equal to \[\dfrac{\sin \theta }{\cos \theta }\]. So, \[x=\dfrac{1}{\cos \theta }+\dfrac{\sin \theta }{\cos \theta }\] or \[x=\dfrac{1+\sin \theta }{\cos \theta }\]. So, we will multiply the expression with \[\cos \theta \] to numerator and denominator we get,

\[x=\dfrac{\left( 1+\sin \theta \right)\times \cos \theta }{{{\cos }^{2}}\theta }\]

Now, we will use the identity, \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \], so we get x as,

\[x=\dfrac{\left( 1+\sin \theta \right)\cos \theta }{{{\cos }^{2}}\theta }\]

Now, we will write \[\left( 1-{{\sin }^{2}}\theta \right)\] as product of \[\left( 1+\sin \theta \right)\] and \[\left( 1-\sin \theta \right)\]. So, we can write x as,

\[x=\dfrac{\left( 1+\sin \theta \right)\cos \theta }{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}\]

Now on simplification we can say that,

\[x=\dfrac{\cos \theta }{1-\sin \theta }\]

So, x is equal to \[\dfrac{\cos \theta }{1-\sin \theta }\].

Now, we will see expression of y,

\[y=\sec \theta -\tan \theta \]

Now, we write \[\sec \theta \] as \[\dfrac{1}{\cos \theta }\] and \[\tan \theta \] as \[\dfrac{\sin \theta }{\cos \theta }\]. So, we can write y as,

\[y=\dfrac{1}{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }=\dfrac{1-\sin \theta }{\cos \theta }\]

We get, \[y=\dfrac{1-\sin \theta }{\cos \theta }\], which can also be written as, \[y=\dfrac{1}{\dfrac{\cos \theta }{1-\sin \theta }}\].

As we know, \[x=\dfrac{\cos \theta }{1-\sin \theta }\]. So, we can write, \[y=\dfrac{1}{x}\].

Here, on cross multiplication we get, \[xy=1\].

Note: There is another method by using the fact that \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\], which can also be written as, \[\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1\]. So, we know that, \[x=\sec \theta +\tan \theta \] and \[y=\sec \theta -\tan \theta \]. Hence, we can say that, \[xy=1\].

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