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. 3
B. 2
C. 1
D. 0

Answer 1

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.