Answer
Verified
480.3k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE