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
•
Total views: 305.7k
•
Views today: 5.83k
Answer
305.7k+ views
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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
