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Question: If \(3\tan \theta \tan \phi =1\), then A. \(\cos \left( \theta +\phi \right)=\cos \left( \theta -...

If 3tan⁡θtan⁡ϕ=13\tan \theta \tan \phi =1, then
A. cos⁡(θ+ϕ)=cos⁡(θ−ϕ)\cos \left( \theta +\phi \right)=\cos \left( \theta -\phi \right)
B. cos⁡(θ+ϕ)=2cos⁡(θ−ϕ)\cos \left( \theta +\phi \right)=2\cos \left( \theta -\phi \right)
C. cos⁡(θ+ϕ)=−cos⁡(θ−ϕ)\cos \left( \theta +\phi \right)=-\cos \left( \theta -\phi \right)
D. 2cos⁡(θ+ϕ)=cos⁡(θ−ϕ)2\cos \left( \theta +\phi \right)=\cos \left( \theta -\phi \right)

Explanation

Solution

Hint : We first use the trigonometric ratio relation of tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}. We add cos⁡θcos⁡ϕ\cos \theta \cos \phi both sides of the equation. We use the associative angle formulas of cos⁡θcos⁡ϕ−sin⁡θsin⁡ϕ=cos⁡(θ+ϕ)\cos \theta \cos \phi -\sin \theta \sin \phi =\cos \left( \theta +\phi \right), sin⁡θsin⁡ϕ+cos⁡θcos⁡ϕ=cos⁡(θ−ϕ)\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 3tan⁡θtan⁡ϕ=13\tan \theta \tan \phi =1 where tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}.
So, 3tan⁡θtan⁡ϕ=3sin⁡θsin⁡ϕcos⁡θcos⁡ϕ=13\tan \theta \tan \phi =3\dfrac{\sin \theta \sin \phi }{\cos \theta \cos \phi }=1.
The cross multiplication gives 3sin⁡θsin⁡ϕ=cos⁡θcos⁡ϕ3\sin \theta \sin \phi =\cos \theta \cos \phi .
Now we add cos⁡θcos⁡ϕ\cos \theta \cos \phi both sides of the equation and get

& 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{aligned}$$ We now apply the associative angle formulas of $$\begin{aligned} & \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{aligned}$$ Therefore, $$\begin{aligned} & \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{aligned}$$ 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.