Question

If $\sin \theta =3\sin \left( \theta +2\alpha \right),$ then the value of $\tan \left( \theta +\alpha \right)+2\tan \alpha$ is:A. 3B. 2C. 1D. 0

Hint: Use the Componendo Dividendo rule in the given expression. Apply trigonometric identities and simplify the expression to get the expression as $\tan \left( \theta +\alpha \right)+2\tan \alpha$.

Complete step by step solution:
Given is the expression $\sin \theta =3\sin \left( \theta +2\alpha \right)$
$\therefore \dfrac{\sin \theta }{\sin \left( \theta +2\alpha \right)}=3.$
Let us use the Componendo Dividendo rule to solve the above expression.
Componendo Dividendo is a theorem on proportions which is used to perform calculations and reduce the number of steps.
According to Componendo Dividendo if $\dfrac{a}{b}=\dfrac{c}{d},$then it implies that $\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}.......(1)$
Thus applying Componendo Dividendo rule in the expression in equation (1),
$\dfrac{\sin \theta }{\sin \left( \theta +2\alpha \right)}=\dfrac{3}{1}......(2)$
Where,
\begin{align} & a=\sin \theta \\ & b=\sin \left( \theta +2\alpha \right) \\ & c=3 \\ & d=1 \\ \end{align}
\begin{align} & \therefore \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d} \\ & \Rightarrow \dfrac{\sin \theta +\sin \left( \theta +2\alpha \right)}{\sin \theta -\sin \left( \theta +2\alpha \right)}=\dfrac{3+1}{3-1} \\ & \Rightarrow \dfrac{\sin \theta +\sin \left( \theta +2\alpha \right)}{\sin \theta -\sin \left( \theta +2\alpha \right)}=\dfrac{4}{2}=2......(3) \\ \end{align}
We know the trigonometric identities,
\begin{align} & \operatorname{sinx}+siny=2sin\left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) \\ & \sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right) \\ \end{align}
Let us apply these identities in equation (3).
$x=\theta$and$y=\left( \theta +2\alpha \right)$.
$\therefore \dfrac{2\sin \left( \dfrac{\theta +\theta +2\alpha }{2} \right)\cos \left( \dfrac{\theta -\theta -2\alpha }{2} \right)}{2\cos \left( \dfrac{\theta +\theta +2\alpha }{2} \right)\sin \left( \dfrac{\theta -\theta -2\alpha }{2} \right)}=2$
By simplifying the expression, we get,
\begin{align} & \Rightarrow \dfrac{\sin \left( \dfrac{2\theta +2\alpha }{2} \right)\cos \left( \dfrac{-2\alpha }{2} \right)}{\cos \left( \dfrac{2\theta +2\alpha }{2} \right)\sin \left( \dfrac{-2\alpha }{2} \right)}=2 \\ & \Rightarrow \dfrac{\sin \left( \theta +\alpha \right)\cos \left( -\alpha \right)}{\cos \left( \theta +\alpha \right)\sin \left( -\alpha \right)}=2 \\ \end{align}
The cosine is an even function, thus $\cos (-\alpha )=\cos \alpha$.
The sine is an odd function, so $sin(-\alpha )=-\sin \alpha$.
$\dfrac{\sin \left( \theta +\alpha \right)\cos \left( \alpha \right)}{-\cos \left( \theta +\alpha \right)\sin \left( \alpha \right)}=2$. By cross multiplying, we get,
$\Rightarrow \dfrac{\sin \left( \theta +\alpha \right)}{\cos \left( \theta +\alpha \right)}=\dfrac{-2\sin \alpha }{\cos \alpha }$.
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }.$
\begin{align} & \therefore \tan (\theta +\alpha )=-2\tan \alpha \\ & \Rightarrow \tan (\theta +\alpha )+2\tan \alpha =0 \\ \end{align}
Thus we got the value of $\tan (\theta +\alpha )+2\tan \alpha$ as 0.
Hence option D is the correct answer.

Note:
Remember the basic trigonometric identities like $(sinA+sinB)$ and $(sinA-sinB)$ which we have used here. They are very important for solving expressions like these. Just apply the formula and simplify it and you will get the answer.