Question

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

Answer 1

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.