
If $3\tan \theta \tan \phi =1$, then
A. $\cos \left( \theta +\phi \right)=\cos \left( \theta -\phi \right)$
B. $\cos \left( \theta +\phi \right)=2\cos \left( \theta -\phi \right)$
C. $\cos \left( \theta +\phi \right)=-\cos \left( \theta -\phi \right)$
D. $2\cos \left( \theta +\phi \right)=\cos \left( \theta -\phi \right)$
Answer
500.4k+ views
Hint: We first use the trigonometric ratio relation of $\tan x=\dfrac{\sin x}{\cos x}$. We add $\cos \theta \cos \phi $ both sides of the equation. We use the associative angle formulas of \[\cos \theta \cos \phi -\sin \theta \sin \phi =\cos \left( \theta +\phi \right)\], \[\sin \theta \sin \phi +\cos \theta \cos \phi =\cos \left( \theta -\phi \right)\] to find the relation between the cos ratios.
Complete step-by-step answer:
We break the given equation $3\tan \theta \tan \phi =1$ where $\tan x=\dfrac{\sin x}{\cos x}$.
So, $3\tan \theta \tan \phi =3\dfrac{\sin \theta \sin \phi }{\cos \theta \cos \phi }=1$.
The cross multiplication gives $3\sin \theta \sin \phi =\cos \theta \cos \phi $.
Now we add $\cos \theta \cos \phi $ both sides of the equation and get
\[\begin{align}
& 3\sin \theta \sin \phi +\cos \theta \cos \phi =\cos \theta \cos \phi +\cos \theta \cos \phi \\
& \Rightarrow 3\sin \theta \sin \phi +\cos \theta \cos \phi =2\cos \theta \cos \phi \\
& \Rightarrow \sin \theta \sin \phi +\cos \theta \cos \phi =2\cos \theta \cos \phi -2\sin \theta \sin \phi \\
\end{align}\]
We now apply the associative angle formulas of
\[\begin{align}
& \cos \theta \cos \phi -\sin \theta \sin \phi =\cos \left( \theta +\phi \right) \\
& \sin \theta \sin \phi +\cos \theta \cos \phi =\cos \left( \theta -\phi \right) \\
\end{align}\]
Therefore,
\[\begin{align}
& \sin \theta \sin \phi +\cos \theta \cos \phi =2\cos \theta \cos \phi -2\sin \theta \sin \phi \\
& \Rightarrow \sin \theta \sin \phi +\cos \theta \cos \phi =2\left( \cos \theta \cos \phi -\sin \theta \sin \phi \right) \\
& \Rightarrow \cos \left( \theta -\phi \right)=2\cos \left( \theta +\phi \right) \\
\end{align}\]
The correct option is D.
So, the correct answer is “Option D”.
Note: The trigonometric equation is tough to assume which thing to add, therefore, it is more likely to first solve the given options in a rough manner to understand the options which give the statement. We then find out the additional expression through back process calculation.
Complete step-by-step answer:
We break the given equation $3\tan \theta \tan \phi =1$ where $\tan x=\dfrac{\sin x}{\cos x}$.
So, $3\tan \theta \tan \phi =3\dfrac{\sin \theta \sin \phi }{\cos \theta \cos \phi }=1$.
The cross multiplication gives $3\sin \theta \sin \phi =\cos \theta \cos \phi $.
Now we add $\cos \theta \cos \phi $ both sides of the equation and get
\[\begin{align}
& 3\sin \theta \sin \phi +\cos \theta \cos \phi =\cos \theta \cos \phi +\cos \theta \cos \phi \\
& \Rightarrow 3\sin \theta \sin \phi +\cos \theta \cos \phi =2\cos \theta \cos \phi \\
& \Rightarrow \sin \theta \sin \phi +\cos \theta \cos \phi =2\cos \theta \cos \phi -2\sin \theta \sin \phi \\
\end{align}\]
We now apply the associative angle formulas of
\[\begin{align}
& \cos \theta \cos \phi -\sin \theta \sin \phi =\cos \left( \theta +\phi \right) \\
& \sin \theta \sin \phi +\cos \theta \cos \phi =\cos \left( \theta -\phi \right) \\
\end{align}\]
Therefore,
\[\begin{align}
& \sin \theta \sin \phi +\cos \theta \cos \phi =2\cos \theta \cos \phi -2\sin \theta \sin \phi \\
& \Rightarrow \sin \theta \sin \phi +\cos \theta \cos \phi =2\left( \cos \theta \cos \phi -\sin \theta \sin \phi \right) \\
& \Rightarrow \cos \left( \theta -\phi \right)=2\cos \left( \theta +\phi \right) \\
\end{align}\]
The correct option is D.
So, the correct answer is “Option D”.
Note: The trigonometric equation is tough to assume which thing to add, therefore, it is more likely to first solve the given options in a rough manner to understand the options which give the statement. We then find out the additional expression through back process calculation.
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