Question

What is \[\cot \left( \dfrac{\theta }{2} \right)\] in terms of trigonometric functions of a unit \[\theta \]?

Answer 1

Hint:This type of question depends on the concept of trigonometry. We use the relation between \[\cot \left( \theta \right)\] and \[\tan \left( \theta \right)\] that is \[\cot \left( \theta \right)=\dfrac{1}{\tan \theta }\]. Also, here we can use the basic definition of \[\tan \theta \]. Also we know that \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] and \[\sin 2\theta =2\sin \theta \cos \theta \].

Complete step by step solution:
Now we have to express \[\cot \left( \dfrac{\theta }{2} \right)\] in terms of trigonometric functions of a unit \[\theta \].
For this let us consider,
\[\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }\]
Let us multiply numerator as well denominator by \[\sin \theta \]
\[\Rightarrow \tan \theta =\dfrac{{{\sin }^{2}}\theta }{\cos \theta \sin \theta }\]
As we know that, \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \] we can write,
\[\Rightarrow \tan \theta =\dfrac{1-{{\cos }^{2}}\theta }{\cos \theta \sin \theta }\]
By multiplying numerator and denominator by 2 we get,
\[\Rightarrow \tan \theta =\dfrac{2-2{{\cos }^{2}}\theta }{2\cos \theta \sin \theta }\]
We know that, \[\sin 2\theta =2\sin \theta \cos \theta \]
\[\Rightarrow \tan \theta =\dfrac{1+1-2{{\cos }^{2}}\theta }{\sin 2\theta }\]
Now we rearrange the numerator
\[\begin{align}
  & \Rightarrow \tan \theta =\dfrac{1-\left( -1 \right)-2{{\cos }^{2}}\theta }{\sin 2\theta } \\
 & \Rightarrow \tan \theta =\dfrac{1-\left( 2{{\cos }^{2}}\theta -1 \right)}{\sin 2\theta } \\
\end{align}\]
As we know that, \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] we can write,
\[\Rightarrow \tan \theta =\dfrac{1-\cos 2\theta }{\sin 2\theta }\]
By replacing \[\theta \] with \[\left( \dfrac{\theta }{2} \right)\] we get,
\[\Rightarrow \tan \left( \dfrac{\theta }{2} \right)=\dfrac{1-\cos \theta }{\sin \theta }\]
Taking reciprocal of both the sides, we get,
\[\Rightarrow \dfrac{1}{\tan \left( \dfrac{\theta }{2} \right)}=\dfrac{1}{\left( \dfrac{1-\cos \theta }{\sin \theta } \right)}\]
By the relation between trigonometric functions we have \[\cot \left( \theta \right)=\dfrac{1}{\tan \theta }\] and hence,
\[\Rightarrow \cot \left( \dfrac{\theta }{2} \right)=\dfrac{\sin \theta }{1-\cos \theta }\]
Thus \[\dfrac{\sin \theta }{1-\cos \theta }\]is the representation of \[\cot \left( \dfrac{\theta }{2} \right)\] in terms of trigonometric functions of a unit \[\theta \].

Note: In this type of question students may make mistakes in understanding the question. Instead of writing in terms of the trigonometric function of a unit \[\theta \] students may try to obtain an expression for \[\theta \]. Also students have to take care in the use of trigonometric formulae also. One of the students may directly obtain an expression for \[\cot \left( \dfrac{\theta }{2} \right)\] by using the basic definition and the same formulae which also gives the same result. Also students have to take care during the conversion from a unit \[\theta \] to half of \[\theta \] that is \[\left( \dfrac{\theta }{2} \right)\].