# 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

Last updated date: 25th Mar 2023

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Answer

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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.

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.

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