Question

If$\tan \theta = n\tan \phi $, then the maximum value of ${\tan ^2}\left( {\theta - \phi } \right)$ is
$\left( a \right){\text{ }}\dfrac{{{{\left( {n - 1} \right)}^2}}}{{4n}}$
$\left( b \right){\text{ }}\dfrac{{\left( {n - 1} \right)}}{{4n}}$
$\left( c \right){\text{ }}\dfrac{{\left( {n - 1} \right)}}{{4{n^2}}}$
$\left( d \right){\text{ }}\dfrac{{{{\left( {n + 1} \right)}^2}}}{{4n}}$

Answer 1

Hint:
First of all we will find them $\tan \left( {\theta - \phi } \right)$by using the formula $\tan \left( {\theta - \phi } \right) = \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}$and then we will differentiate the equation ${\tan ^2}\left( {\theta - \phi } \right)$and from here we will get the value$\tan \phi $. And in this way, we can get the value.

Formula used:
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
And also the derivative for the division will be
$\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$

Complete step by step solution:
As from the formula $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
We can write according to the question as
$\tan \left( {\theta - \phi } \right) = \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}$
And also it is given that $\tan \theta = n\tan \phi $
And from here it can be written as
$ \Rightarrow \dfrac{{\left( {n - 1} \right)\tan \phi }}{{1 + n{{\tan }^2}\phi }}$
Let suppose ${\tan ^2}\left( {\theta - \phi } \right)$be $y$
Therefore,
$y = {\tan ^2}\left( {\theta - \phi } \right)$
Now putting the value from the above we get
$ \Rightarrow y = \dfrac{{{{\left( {n - 1} \right)}^2}{{\tan }^2}\phi }}{{{{\left( {1 + n{{\tan }^2}\phi } \right)}^2}}}$
Now differentiating it with respect to $\phi $
$ \Rightarrow \dfrac{{dy}}{{d\phi }} = \dfrac{{{{\left( {1 + n{{\tan }^2}\phi } \right)}^2}\left[ {{{\left( {n - 1} \right)}^2}2\tan \phi .{{\sec }^2}\phi } \right] - {{\left( {n - 1} \right)}^2}{{\tan }^2}\phi \left[ {2\left( {1 + n{{\tan }^2}\phi } \right)2.n.\tan \phi .{{\sec }^2}\phi } \right]}}{{{{\left( {1 + n{{\tan }^2}\phi } \right)}^4}}}$
So here,
$\dfrac{{dy}}{{d\phi }} = 0$, for maximum.
So on equating, we get
$0 \Rightarrow 2{\left( {n - 1} \right)^2}\tan \phi .{\sec ^2}\phi \left( {1 + n{{\tan }^2}\phi } \right)\left[ {1 + n{{\tan }^2}\phi - 2n{{\tan }^2}\phi } \right]$
On further solving more, we get
$0 \Rightarrow 2{\left( {n - 1} \right)^2}\tan \phi .{\sec ^2}\phi \left( {1 + n{{\tan }^2}\phi } \right)\left[ {1 - n{{\tan }^2}\phi } \right]$
So now on equating each term with the LHS, we get
$ \Rightarrow {\tan ^2}\phi = 0$, and also $1 - n{\tan ^2}\phi = 0$
And from here ${\tan ^2}\phi = \dfrac{1}{n}$
So now putting the above value of${\tan ^2}\phi $in the value of$y$, we get
$ \Rightarrow y = \dfrac{{{{\left( {n - 1} \right)}^2}0}}{{{{\left( {1 + n0} \right)}^2}}} = 0$
And for the other value of${\tan ^2}\phi $, we get
$ \Rightarrow y = \dfrac{{{{\left( {n - 1} \right)}^2}\dfrac{1}{n}}}{{{{\left( {1 + n\dfrac{1}{n}} \right)}^2}}}$
And on further solving more, we get
$ \Rightarrow y = \dfrac{{{{\left( {n - 1} \right)}^2}}}{4}\dfrac{1}{n}$
And it will be equal to
$ \Rightarrow y = \dfrac{{{{\left( {n - 1} \right)}^2}}}{{4n}}$
Therefore, $y = \dfrac{{{{\left( {n - 1} \right)}^2}}}{{4n}}$will be the maximum value of${\tan ^2}\left( {\theta - \phi } \right)$.

Hence, the option $\left( a \right)$will be correct.

Note:
As we have seen that without formula and identities we can find the values of such functions. To find the appropriate formula at one click would come through the experience gained and the more basic formulas we have memorized. Also, we should always try to use sin and cos form because most of the formulas are based on this. It can be easily solved then. So we should always try to learn all the formulas related to trigonometric to solve the problem without any error and easily.