If a trigonometric expression is given as \[\sin \theta =n\sin \left( \theta +2\alpha \right)\] then \[\tan \left( \theta +\alpha \right)=\]
\[\left( a \right)\dfrac{1+n}{1-n}\tan \alpha \]
\[\left( b \right)\dfrac{1-n}{1+n}\tan \alpha \]
\[\left( c \right)\tan \alpha \]
\[\left( d \right)\text{None}\]

Hint: We are asked to find the value of \[\tan \left( \theta +\alpha \right),\] we start by simplifying \[\sin \theta =n\sin \left( \theta +2\alpha \right)\] as \[\dfrac{\sin \theta }{\sin \left( \theta +2\alpha \right)}=n.\] Then we can apply componendo and dividendo which says if \[\dfrac{a}{b}=\dfrac{c}{d}\] then \[\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}.\] Then using \[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] and \[\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right).\] We will simplify our terms at last. Then we will use \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] to get our required solution.

Complete step-by-step solution:
We have been given in the question that \[\sin \theta =n\sin \left( \theta +2\alpha \right).\] On simplifying we get, \[\dfrac{\sin \theta }{\sin \left( \theta +2\alpha \right)}=n.\]
Now we will use componendo and dividendo rule which says that if \[\dfrac{a}{b}=\dfrac{c}{d}\] then \[\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}.\] So, we apply this on our equation we got above, i.e. \[\dfrac{\sin \theta }{\sin \left( \theta +2\alpha \right)}=\dfrac{n}{1}.\] So, we get,
\[\dfrac{\sin \theta +\sin \left( \theta +2a \right)}{\sin \theta -\sin \left( \theta +2\alpha \right)}=\dfrac{n+1}{n-1}\]
Now, we know the trigonometric formulas, i.e.
\[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]
\[\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)\]
So, take A as \[\theta \] and B as \[\theta +2\alpha .\] So, we get,
\[\dfrac{n+1}{n-1}=\dfrac{2\sin \left( \dfrac{\theta +\theta +2\alpha }{2} \right)\cos \left( \dfrac{\theta -\left( \theta +2\alpha \right)}{2} \right)}{2\cos \left( \dfrac{\theta +\theta +2\alpha }{2} \right)\sin \left( \dfrac{\theta -\left( \theta +2\alpha \right)}{2} \right)}\]
Simplifying further, we get,
\[\Rightarrow \dfrac{2\sin \left( \theta +\alpha \right)\cos \left( -\alpha \right)}{2\sin \left( -\alpha \right)\cos \left( \theta +\alpha \right)}=\dfrac{n+1}{n-1}\]
Now we know that \[\cos \left( -\theta \right)=\cos \theta \] and \[\sin \left( -\theta \right)=-\sin \theta .\] So, we get,
\[\Rightarrow \dfrac{2\sin \left( \theta +\alpha \right)\cos \alpha }{-2\sin \alpha \cos \left( \theta +\alpha \right) }=\dfrac{n+1}{n-1}\]
Now, separating the terms, we get,
\[\Rightarrow \dfrac{-\sin \left( \theta +\alpha \right)}{\cos \left( \theta +\alpha \right) }.\dfrac{\cos \alpha }{\sin \alpha }=\dfrac{n+1}{n-1}\]
As, \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta ,\] we get,
\[\Rightarrow \dfrac{-\tan \left( \theta +\alpha \right)}{\tan \alpha }=\dfrac{n+1}{n-1}\]
On shifting \[\tan \alpha \] we get,
\[\Rightarrow -\tan \left( \theta +\alpha \right) =\dfrac{n+1}{n-1}\tan \alpha \]
On multiplying both the sides, we get,
\[\Rightarrow \tan \left( \theta +\alpha \right)=\dfrac{-\left( n+1 \right)}{\left( n-1 \right)}\tan a\]
Now, $- (n – 1) = 1 – n$. So, we get,
\[\Rightarrow \tan \left( \theta +\alpha \right)=\dfrac{n+1}{1-n}\tan \alpha \]
Or it can be written as,
\[\Rightarrow \tan \left( \theta +\alpha \right)=\dfrac{1+n}{1-n}\tan \alpha \]
Hence, the right option is (a).

Note: Remember that \[\dfrac{\theta -\left( \theta +2\alpha \right)}{2}\] will give us \[\dfrac{\theta -\theta -2\alpha }{2}=\dfrac{0-2\alpha }{2}=-\alpha \] and \[\dfrac{\theta +\left( \theta +2\alpha \right)}{2}\] will give us \[\dfrac{\theta +\theta +2\alpha }{2}=\dfrac{2\theta +2\alpha }{2}=\theta +\alpha .\] Also, \[\dfrac{\cos \alpha }{\sin \alpha }=\dfrac{1}{\dfrac{\sin \alpha }{\cos \alpha }}\] as \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta .\] So, \[\dfrac{1}{\dfrac{\sin \alpha }{\cos \alpha }}=\dfrac{1}{\tan \alpha }.\] Lastly, always remember that the product of two negatives is always positive.