# 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.

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.

Last updated date: 03rd Oct 2023

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