Question

# Given that, $x=\sec \theta +\tan \theta ,y=\sec \theta -\tan$. Establish a relation between x and y by eliminating ‘$\theta$’.

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 $\cos \theta$.

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, $xy=1$.