If \[\cos ec\theta = \dfrac{{p + q}}{{p - q}}\], then \[\cot \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right) = \]
A) \[\sqrt {\dfrac{q}{p}} \]
B) \[\sqrt {\dfrac{p}{q}} \]
C) \[pq\]
D) \[\sqrt {pq} \]

Hint: Here we will first use the following identity:-
\[\cos ec\theta = \dfrac{1}{{\sin \theta }}\] then we will use componendo dividendo and then finally use the following identities to get the desired answer.
\[
  \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} \\
  \cot \theta = \dfrac{1}{{\tan \theta }} \\
 \]

Complete step-by-step answer:
The given equation is:-
\[\cos ec\theta = \dfrac{{p + q}}{{p - q}}\]
Now applying the following identity
\[\cos ec\theta = \dfrac{1}{{\sin \theta }}\]
We get:-
\[\dfrac{1}{{\sin \theta }} = \dfrac{{p + q}}{{p - q}}\]
Now applying componendo and dividendo we get:-
\[\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} = \dfrac{{p + q + p - q}}{{p + q - \left( {p - q} \right)}}\]
Solving it further we get:-
\[
  \dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} = \dfrac{{2p}}{{p + q - p + q}} \\
   \Rightarrow \dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} = \dfrac{{2p}}{{2q}} \\
   \Rightarrow \dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} = \dfrac{p}{q} \\
 \]
Now we know that:-
\[\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}\]
Hence substituting the value we get:-
\[\dfrac{{1 + 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{1 - 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}} = \dfrac{p}{q}\]……………………………….(1)
Now we know that:-
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Hence, \[{\sin ^2}\dfrac{\theta }{2} + {\cos ^2}\dfrac{\theta }{2} = 1\]
Hence substituting this value in equation1 we get:-
\[\dfrac{{{{\sin }^2}\dfrac{\theta }{2} + {{\cos }^2}\dfrac{\theta }{2} + 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{{{\sin }^2}\dfrac{\theta }{2} + {{\cos }^2}\dfrac{\theta }{2} - 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}} = \dfrac{p}{q}\]
Now using the following identities:-
\[
  {\left( {A + B} \right)^2} = {A^2} + {B^2} + 2AB \\
  {\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB \\
 \]
We get:-
\[\dfrac{{{{\left( {\sin \dfrac{\theta }{2} + \cos \dfrac{\theta }{2}} \right)}^2}}}{{{{\left( {\sin \dfrac{\theta }{2} - \cos \dfrac{\theta }{2}} \right)}^2}}} = \dfrac{p}{q}\]
Solving it further we get:-
\[\dfrac{{{{\left( {\sin \dfrac{\theta }{2} + \cos \dfrac{\theta }{2}} \right)}^2}}}{{{{\left( {\cos \dfrac{\theta }{2} - \sin \dfrac{\theta }{2}} \right)}^2}}} = \dfrac{p}{q}\]
Now taking \[{\cos ^2}\dfrac{\theta }{2}\] common from numerator and denominator we get:-
\[{\left( {\dfrac{{\dfrac{{\cos \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}} + \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}}}{{\dfrac{{\cos \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}} - \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}}}} \right)^2} = \dfrac{p}{q}\]
Now we know that:
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Hence, \[\tan \dfrac{\theta }{2} = \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}\]
Therefore substituting the values we get:-
\[{\left( {\dfrac{{1 + \tan \dfrac{\theta }{2}}}{{1 - \tan \dfrac{\theta }{2}}}} \right)^2} = \dfrac{p}{q}\]
Now we know that
\[\tan \dfrac{\pi }{4} = 1\]
Hence substituting the value we get:-
\[{\left( {\dfrac{{\tan \dfrac{\pi }{4} + \tan \dfrac{\theta }{2}}}{{\tan \dfrac{\pi }{4} - \tan \dfrac{\theta }{2}}}} \right)^2} = \dfrac{p}{q}\]
Now we know that
\[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
Hence applying this identity in above equation we get:-
\[{\left[ {\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right)} \right]^2} = \dfrac{p}{q}\]
Now taking square root of both the sides we get:-
\[\sqrt {{{\left[ {\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right)} \right]}^2}} = \sqrt {\dfrac{p}{q}} \]
Simplifying it further we get:-
\[\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right) = \sqrt {\dfrac{p}{q}} \]……………………….(2)
Now we know that
\[\cot \theta = \dfrac{1}{{\tan \theta }}\]
Hence we will take reciprocal of equation 2 and then apply this identity to get desired value:-
\[\dfrac{1}{{\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right)}} = \sqrt {\dfrac{q}{p}} \]
Now applying the identity we get:-
\[\cot \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right) = \sqrt {\dfrac{q}{p}} \]
Therefore option B is the correct option.

Note: Students might make mistake in making the squares of the quantities using the identities:
\[
  {\left( {A + B} \right)^2} = {A^2} + {B^2} + 2AB \\
  {\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB \\
 \]
In such questions we need to use the given information and then then transform it into required form to get the desired answer.