Question

# 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.
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} \\ \cot \theta = \dfrac{1}{{\tan \theta }} \\$

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.