How do you simplify \[\tan \left( \dfrac{\theta }{2} \right)\] using the double angle identities?

Hint: In order to simplify the given function using the double angle identity, first by using the relation between the sine, cosine and tangent function i.e. \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] . Rewrite the given function in the form of sine and cosine function. Later using the double angles formula of sine, cosine function and solving the expression by substituting the values, we will get the exact values of \[\tan \left( \dfrac{\theta }{2} \right)\] .
Formula used:
 \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]
 \[\sin 2x=2\sin x\cos x\]
 \[\cos 2x=2{{\cos }^{2}}x-1\]

Complete step-by-step answer:
We have given that,
 \[\Rightarrow \tan \left( \dfrac{\theta }{2} \right)\]
As we know that,
The relation between the trigonometric function of sine, cosine and tangent function is given by;
 \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]
Therefore,
 \[\Rightarrow \tan \left( \dfrac{\theta }{2} \right)=\dfrac{\sin \left( \dfrac{\theta }{2} \right)}{\cos \left( \dfrac{\theta }{2} \right)}\]
Using the double angle identities of sine and cosine function of trigonometry is given by;
 \[\sin 2x=2\sin x\cos x\]
And
 \[\cos 2x=2{{\cos }^{2}}x-1\]
 \[\Rightarrow \dfrac{\sin \left( \dfrac{\theta }{2} \right)}{\cos \left( \dfrac{\theta }{2} \right)}=\dfrac{2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)}{2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)}\]
Now,
Solving: \[2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)\]
Using the double angle identity of cosine function is given by,
 \[\cos 2x=2{{\cos }^{2}}x-1\]
In the given above question,
Let \[\dfrac{\theta }{2}=x\]
Therefore,
Applying the double angle identities;
Substituting \[\dfrac{\theta }{2}=x\] , we will get
 \[\cos \left( 2\times \left( \dfrac{\theta }{2} \right) \right)=2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)-1\]
Adding 1 to both the sides,
 \[\cos \left( 2\times \left( \dfrac{\theta }{2} \right) \right)+1=2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)-1+1\]
Simplifying the above, we will get
 \[\Rightarrow 1+\cos \theta =2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)\] ------- (1)
Solving: \[2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)\]
Now,
We have,
 \[\Rightarrow 2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)\]
We can rewrite it as,
 \[\Rightarrow 2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)=\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)+\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)\]
As we know that,
 \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\]
Applying this identity of sine function,
Here \[A=\left( \dfrac{\theta }{2} \right)\ ,\ B=\left( \dfrac{\theta }{2} \right)\]
 \[\Rightarrow 2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)=\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)+\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)=\sin \left( \left( \dfrac{\theta }{2} \right)+\left( \dfrac{\theta }{2} \right) \right)\]
 \[\Rightarrow 2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)=\sin \left( \left( \dfrac{\theta }{2} \right)+\left( \dfrac{\theta }{2} \right) \right)\]
Solving the above,
 \[\Rightarrow 2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)=\sin \left( \dfrac{2\theta }{2} \right)=\sin \theta \]
Therefore,
 \[\Rightarrow 2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)=\sin \theta \] ------- (2)
Substituting the value from equation (1) and equation (2), we will get
 \[\Rightarrow \tan \left( \dfrac{\theta }{2} \right)=\dfrac{\sin \theta }{1+\cos \theta }\]
Hence, this is the required answer.
So, the correct answer is “ \[\dfrac{\sin \theta }{1+\cos \theta }\] ”.

Note: While solving these types of questions, students must need to remember the formulas of double angle of trigonometric functions. One must be careful while noted down the values to avoid any error in the answer. They also need to consider the quadrant which is given in the question and they must remember which trigonometric function is positive or negative in all the four quadrants. We must know the basic concept of trigonometry and also the definition of the trigonometric function. The sine, cosine and the tangent are the three basic functions in introduction to trigonometry which shows the relation between all the sides of the triangles.