If $x = \sec \theta - \tan \theta $ , $y = \cos ec\theta + \cot \theta $ , then
(A) $x = \dfrac{{\left( {y + 1} \right)}}{{\left( {y - 1} \right)}}$
(B) $x = \dfrac{{\left( {y - 1} \right)}}{{\left( {y + 1} \right)}}$
(C) $y = \dfrac{{\left( {1 - x} \right)}}{{\left( {1 + x} \right)}}$
(D) None of these

Hint: While solving such questions mostly we first need to convert $\tan \theta ,\sec \theta ,\cos ec\theta $ and $\cot \theta $ into $\sin \theta $ and $\cos \theta $ . In the given question we will use this approach and find x and y in terms of $\sin \theta $ and $\cos \theta $. After this, we solve either x in terms of y or y in terms of x to get a relation between them.

Formula Used:
$\cos 2\theta = 2{\cos ^2}\theta - 1$ , $\sin 2\theta = 2\sin \theta \cos \theta $ , $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta $ , ${\sin ^2}\left( {\dfrac{\theta }{2}} \right) + {\cos ^2}\left( {\dfrac{\theta }{2}} \right) = 1$ , \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] , ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$

Complete step by step Solution:
Given: $x = \sec \theta - \tan \theta $
And $y = \cos ec\theta + \cot \theta $
Converting x and y in terms of $\sin \theta $ and $\cos \theta $ :
 $x = \sec \theta - \tan \theta $
 $x = \dfrac{1}{{\cos \theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}$
We get,
 $x = \dfrac{{1 - \sin \theta }}{{\cos \theta }}$ ...(1)
Now we will solve for y as follows,
 $y = \cos ec\theta + \cot \theta $
 $y = \dfrac{1}{{\sin \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}$
 We get,
 $y = \dfrac{{1 + \cos \theta }}{{\sin \theta }}$ ...(2)
Now we will use half-angle formulae in trigonometry.
We know that $\cos 2\theta = 2{\cos ^2}\theta - 1$
And $\sin 2\theta = 2\sin \theta \cos \theta $
Using them in equation (2) we get,
 $y = \dfrac{{2{{\cos }^2}\left( {\dfrac{\theta }{2}} \right)}}{{2\sin \left( {\dfrac{\theta }{2}} \right)\cos \left( {\dfrac{\theta }{2}} \right)}}$
 $y = \dfrac{{\cos \left( {\dfrac{\theta }{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}$
And we know that $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta $
Thus, we get $y = \cot \left( {\dfrac{\theta }{2}} \right)$ ...(3)
Now we will solve x for y.
Now since y has term $\left( {\dfrac{\theta }{2}} \right)$ therefore we will use half angle formulae to solve x further.
From equation (1) we got, $x = \dfrac{{1 - \sin \theta }}{{\cos \theta }}$
This implies that \[x = \dfrac{{1 - 2\sin \left( {\dfrac{\theta }{2}} \right)\cos \left( {\dfrac{\theta }{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{\theta }{2}} \right) - {{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}\] ...(4)
This is because $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $
Also, substitute ${\sin ^2}\left( {\dfrac{\theta }{2}} \right) + {\cos ^2}\left( {\dfrac{\theta }{2}} \right) = 1$ in equation (4), we get
 \[x = \dfrac{{{{\sin }^2}\left( {\dfrac{\theta }{2}} \right) + {{\cos }^2}\left( {\dfrac{\theta }{2}} \right) - 2\sin \left( {\dfrac{\theta }{2}} \right)\cos \left( {\dfrac{\theta }{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{\theta }{2}} \right) - {{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}\]
Now we will use the identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] in numerator.
 \[x = \dfrac{{{{\left( {\sin \left( {\dfrac{\theta }{2}} \right) - \cos \left( {\dfrac{\theta }{2}} \right)} \right)}^2}}}{{{{\cos }^2}\left( {\dfrac{\theta }{2}} \right) - {{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}\]
Divide both numerator and denominator by ${\sin ^2}\left( {\dfrac{\theta }{2}} \right)$ so that we can get x in terms of $\cot \left( {\dfrac{\theta }{2}} \right)$ and hence in terms of y.
Therefore, we get
  \[x = \dfrac{{{{\left( {1 - \cot \left( {\dfrac{\theta }{2}} \right)} \right)}^2}}}{{{{\cot }^2}\left( {\dfrac{\theta }{2}} \right) - 1}}\]
Now substitute $y = \cot \left( {\dfrac{\theta }{2}} \right)$ from equation (2).
We get, \[x = \dfrac{{{{\left( {1 - y} \right)}^2}}}{{{y^2} - 1}}\]
Use the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator, we get
 \[x = \dfrac{{{{\left( {y - 1} \right)}^2}}}{{\left( {y + 1} \right)\left( {y - 1} \right)}}\]
Thus we get \[x = \dfrac{{\left( {y - 1} \right)}}{{\left( {y + 1} \right)}}\] which is option B.

Hence, the correct option is (B).

Note:Whenever we face such questions we should try to first simplify them in terms of $\sin \theta $ and $\cos \theta $ and then solve the rest though this is not a rule and is not compulsory but makes it easier to solve the question. Trigonometric identities should be used carefully as the same identity can be written in different ways at different times.