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Question: If \[\cos ec\theta = \dfrac{{p + q}}{{p - q}}\], then \[\cot \left( {\dfrac{\pi }{4} + \dfrac{\theta...

If cos⁡ecθ=p+qp−q\cos ec\theta = \dfrac{{p + q}}{{p - q}}, then cot⁡(π4+θ2)=\cot \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right) =
A) qp\sqrt {\dfrac{q}{p}}
B) pq\sqrt {\dfrac{p}{q}}
C) pqpq
D) pq\sqrt {pq}

Explanation

Solution

Here we will first use the following identity:-
cos⁡ecθ=1sin⁡θ\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⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B cot⁡θ=1tan⁡θ  \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⁡ecθ=p+qp−q\cos ec\theta = \dfrac{{p + q}}{{p - q}}
Now applying the following identity
cos⁡ecθ=1sin⁡θ\cos ec\theta = \dfrac{1}{{\sin \theta }}
We get:-
1sin⁡θ=p+qp−q\dfrac{1}{{\sin \theta }} = \dfrac{{p + q}}{{p - q}}
Now applying componendo and dividendo we get:-
1+sin⁡θ1−sin⁡θ=p+q+p−qp+q−(p−q)\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} = \dfrac{{p + q + p - q}}{{p + q - \left( {p - q} \right)}}
Solving it further we get:-

1+sin⁡θ1−sin⁡θ=2pp+q−p+q ⇒1+sin⁡θ1−sin⁡θ=2p2q ⇒1+sin⁡θ1−sin⁡θ=pq  \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⁡θ=2sin⁡θ2cos⁡θ2\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}
Hence substituting the value we get:-
1+2sin⁡θ2cos⁡θ21−2sin⁡θ2cos⁡θ2=pq\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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Hence, sin⁡2θ2+cos⁡2θ2=1{\sin ^2}\dfrac{\theta }{2} + {\cos ^2}\dfrac{\theta }{2} = 1
Hence substituting this value in equation1 we get:-
sin⁡2θ2+cos⁡2θ2+2sin⁡θ2cos⁡θ2sin⁡2θ2+cos⁡2θ2−2sin⁡θ2cos⁡θ2=pq\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:-

(A+B)2=A2+B2+2AB (A−B)2=A2+B2−2AB  {\left( {A + B} \right)^2} = {A^2} + {B^2} + 2AB \\\ {\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB \\\

We get:-
(sin⁡θ2+cos⁡θ2)2(sin⁡θ2−cos⁡θ2)2=pq\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:-
(sin⁡θ2+cos⁡θ2)2(cos⁡θ2−sin⁡θ2)2=pq\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⁡2θ2{\cos ^2}\dfrac{\theta }{2} common from numerator and denominator we get:-
(cos⁡θ2cos⁡θ2+sin⁡θ2cos⁡θ2cos⁡θ2cos⁡θ2−sin⁡θ2cos⁡θ2)2=pq{\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⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
Hence, tan⁡θ2=sin⁡θ2cos⁡θ2\tan \dfrac{\theta }{2} = \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}
Therefore substituting the values we get:-
(1+tan⁡θ21−tan⁡θ2)2=pq{\left( {\dfrac{{1 + \tan \dfrac{\theta }{2}}}{{1 - \tan \dfrac{\theta }{2}}}} \right)^2} = \dfrac{p}{q}
Now we know that
tan⁡π4=1\tan \dfrac{\pi }{4} = 1
Hence substituting the value we get:-
(tan⁡π4+tan⁡θ2tan⁡π4−tan⁡θ2)2=pq{\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⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}
Hence applying this identity in above equation we get:-
[tan⁡(π4+θ2)]2=pq{\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:-
[tan⁡(π4+θ2)]2=pq\sqrt {{{\left[ {\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right)} \right]}^2}} = \sqrt {\dfrac{p}{q}}
Simplifying it further we get:-
tan⁡(π4+θ2)=pq\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right) = \sqrt {\dfrac{p}{q}} ……………………….(2)
Now we know that
cot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }}
Hence we will take reciprocal of equation 2 and then apply this identity to get desired value:-
1tan⁡(π4+θ2)=qp\dfrac{1}{{\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right)}} = \sqrt {\dfrac{q}{p}}
Now applying the identity we get:-
cot⁡(π4+θ2)=qp\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:

(A+B)2=A2+B2+2AB (A−B)2=A2+B2−2AB  {\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.