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If the trigonometric equations given as ${{\tan }^{3}}\theta =\tan \phi $ and $\tan 2\theta =2\tan \alpha $, prove that $\theta +\phi =n\pi +\alpha $.

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Hint: Use the trigonometric identity related to tan that is $\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$ . Use the given condition to find the value of $\theta +\phi $ and find out the required relation.
As per the given conditions, we have
${{\tan }^{3}}\theta =\tan \phi ..........(i)$
$\tan 2\theta =2\tan \alpha ........(ii)$
Now consider the expression,
$\tan (\theta +\phi )$
Now we know, as per the trigonometric identity, $\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$, so the above expression can be written as,
$\tan (\theta +\phi )=\dfrac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }$
Now substituting the value from equation (i), we get
$\tan (\theta +\phi )=\dfrac{\tan \theta +{{\tan }^{3}}\theta }{1-\tan \theta {{\tan }^{3}}\theta }$
Taking out the common term from the numerator, we get
$\tan (\theta +\phi )=\dfrac{\tan \theta (1+{{\tan }^{2}}\theta )}{1-{{\tan }^{4}}\theta }$
Now we know, ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$, using this identity the above expression can be written as,
$\tan (\theta +\phi )=\dfrac{\tan \theta (1+{{\tan }^{2}}\theta )}{\left( 1-{{\tan }^{2}}\theta \right)\left( 1+{{\tan }^{2}}\theta \right)}$
Cancelling the like terms, we get
$\tan (\theta +\phi )=\dfrac{\tan \theta }{\left( 1-{{\tan }^{2}}\theta \right)}$
Multiplying and dividing by ‘2’, we get
$\tan (\theta +\phi )=\dfrac{2\tan \theta }{\left( 1-{{\tan }^{2}}\theta \right)}\times \dfrac{1}{2}$
Now we know, $\dfrac{2\tan \theta }{\left( 1-{{\tan }^{2}}\theta \right)}=\tan 2\theta $, substituting this in above expression, we get
$\tan (\theta +\phi )=\tan 2\theta \times \dfrac{1}{2}$
Substituting value from equation (ii), we get
$\tan (\theta +\phi )=2\tan \alpha \times \dfrac{1}{2}$
Cancelling the like terms, we get
$\tan (\theta +\phi )=\tan \alpha ..........(iii)$
Now consider the expression,
$\tan (n\pi +\alpha )$
Now we know, as per the trigonometric identity, $\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$, so the above expression can be written as,
$\tan (n\pi +\alpha )=\dfrac{\tan n\pi +\tan \alpha }{1-\tan n\pi \tan \alpha }$
Now we know, $\tan n\pi =0$, substituting this value in above expression, we get
$\tan (n\pi +\alpha )=\dfrac{0+\tan \alpha }{1-(0)\tan \alpha }=\tan \alpha $
Substituting this value in equation (iii), we get
$\begin{align}
  & \tan (\theta +\phi )=\tan \left( n\pi +\alpha \right) \\
 & \Rightarrow \theta +\phi =n\pi +\alpha \\
\end{align}$
Hence proved
Note: Sometimes students get confused after the step $\tan (\theta +\phi )=\tan \alpha $ if they don’t remember $\tan n\pi =0$.
So it’s very important to remember important identities.
One more approach is to find the value of $\tan (\theta +\phi )$ in terms of $\theta $. Next find the value of $\tan \left( n\pi +\alpha \right)$ in terms of $\theta $ and then equate.
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