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Question: Given that, \[x=\sec \theta +\tan \theta ,y=\sec \theta -\tan \]. Establish a relation between x and...

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

Explanation

Solution

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

Complete step-by-step answer:
In the question we are given expressions of x and y in terms of θ\theta and we are asked to form an equation or relation between x and y by elimination θ\theta .
We are given that,
x=sec⁡θ+tan⁡θx=\sec \theta +\tan \theta
We know that sec⁡θ\sec \theta is equal to 1cos⁡θ\dfrac{1}{\cos \theta } and tan⁡θ\tan \theta is equal to sin⁡θcos⁡θ\dfrac{\sin \theta }{\cos \theta }. So, x=1cos⁡θ+sin⁡θcos⁡θx=\dfrac{1}{\cos \theta }+\dfrac{\sin \theta }{\cos \theta } or x=1+sin⁡θcos⁡θx=\dfrac{1+\sin \theta }{\cos \theta }. So, we will multiply the expression with cos⁡θ\cos \theta to numerator and denominator we get,
x=(1+sin⁡θ)×cos⁡θcos⁡2θx=\dfrac{\left( 1+\sin \theta \right)\times \cos \theta }{{{\cos }^{2}}\theta }
Now, we will use the identity, cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta , so we get x as,
x=(1+sin⁡θ)cos⁡θcos⁡2θx=\dfrac{\left( 1+\sin \theta \right)\cos \theta }{{{\cos }^{2}}\theta }
Now, we will write (1−sin⁡2θ)\left( 1-{{\sin }^{2}}\theta \right) as product of (1+sin⁡θ)\left( 1+\sin \theta \right) and (1−sin⁡θ)\left( 1-\sin \theta \right). So, we can write x as,
x=(1+sin⁡θ)cos⁡θ(1+sin⁡θ)(1−sin⁡θ)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=cos⁡θ1−sin⁡θx=\dfrac{\cos \theta }{1-\sin \theta }
So, x is equal to cos⁡θ1−sin⁡θ\dfrac{\cos \theta }{1-\sin \theta }.
Now, we will see expression of y,
y=sec⁡θ−tan⁡θy=\sec \theta -\tan \theta
Now, we write sec⁡θ\sec \theta as 1cos⁡θ\dfrac{1}{\cos \theta } and tan⁡θ\tan \theta as sin⁡θcos⁡θ\dfrac{\sin \theta }{\cos \theta }. So, we can write y as,
y=1cos⁡θ−sin⁡θcos⁡θ=1−sin⁡θcos⁡θy=\dfrac{1}{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }=\dfrac{1-\sin \theta }{\cos \theta }
We get, y=1−sin⁡θcos⁡θy=\dfrac{1-\sin \theta }{\cos \theta }, which can also be written as, y=1cos⁡θ1−sin⁡θy=\dfrac{1}{\dfrac{\cos \theta }{1-\sin \theta }}.
As we know, x=cos⁡θ1−sin⁡θx=\dfrac{\cos \theta }{1-\sin \theta }. So, we can write, y=1xy=\dfrac{1}{x}.
Here, on cross multiplication we get, xy=1xy=1.

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