Question

Let’s assume the values of $u,v$ are $${\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$$, $${\tan ^{ - 1}}\left( x \right)$$ . Now choose the value of $\dfrac{{du}}{{dv}}$ from the options given below.
1. $2$
2. $1$
3. $\dfrac{1}{2}$
4. $$ - 1$$

Answer 1

Hint: There are various ways of solving this problem. Now we want the value of $\dfrac{{du}}{{dv}}$ so let’s try to express the value of u in terms of v to calculate the value of $\dfrac{{du}}{{dv}}$ and then we can easily calculate the value of $\dfrac{{du}}{{dv}}$ by using the trigonometric relations we know.

Complete step-by-step solution:
Given,
$u = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$,
$v = {\tan ^{ - 1}}\left( x \right)$
On applying $\tan $a function on both sides we get
$\tan v = x$.
Now, let’s substitute the value $x$ in $u$. We get,
$u = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {{\tan }^2}v} - 1}}{{\tan v}}} \right)$
We know that
$\tan \left( {\dfrac{v}{2}} \right) = \dfrac{{\sqrt {1 + {{\tan }^2}v} - 1}}{{\tan v}}$
Now let’s replace the value of $\dfrac{{\sqrt {1 + {{\tan }^2}v} - 1}}{{\tan v}}$ in $u$ into $\tan \left( {\dfrac{v}{2}} \right)$.
Therefore,$u = {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{v}{2}} \right)} \right)$
We know that the value of ${\tan ^{ - 1}}\left( {\tan \left( x \right)} \right)$ is nothing but $x$. Since $\tan $ and ${\tan ^{ - 1}}$ are inverse functions.
So, $u = \dfrac{v}{2}$
Now, let’s calculate the value of $\dfrac{{du}}{{dv}}$,
By substituting the value of $u$ in terms of $v$ we get $\dfrac{{du}}{{dv}} = \dfrac{d}{{dv}}\left( {\dfrac{v}{2}} \right)$,
$ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{1}{2}\dfrac{d}{{dv}}\left( v \right)$
$ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{1}{2}\dfrac{{dv}}{{dv}}$
$ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{1}{2}$
So, the required value $\dfrac{{du}}{{dv}}$ is $\dfrac{1}{2}$.
The correct option is 3.
Additional information:
In the above solution, we have used the formula of $\tan \left( {\dfrac{v}{2}} \right)$.
Now let’s prove it.
The formula is
$\tan \left( {\dfrac{v}{2}} \right) = \dfrac{{\sqrt {1 + {{\tan }^2}v} - 1}}{{\tan v}}$,
$RHS = \dfrac{{\sqrt {1 + {{\tan }^2}v} - 1}}{{\tan v}}$,
We know by the basic trigonometric equations ${\sec ^2}\theta - {\tan ^2}\theta = 1$$ \Leftrightarrow $${\sec ^2}\theta = 1 + {\tan ^2}\theta $
Therefore,
$ \Rightarrow RHS = \dfrac{{\sqrt {{{\sec }^2}v} - 1}}{{\tan v}}$,
$ \Rightarrow RHS = \dfrac{{\sec v - 1}}{{\tan v}}$,
On multiplying both numerator and denominator by $\cos v$, we get
$ \Rightarrow RHS = \dfrac{{1 - \cos v}}{{\sin v}}$,
We know that,
$1 - \cos v = 2{\sin ^2}\left( {\dfrac{v}{2}} \right),\sin v = 2\sin \left( {\dfrac{v}{2}} \right)\cos \left( {\dfrac{v}{2}} \right)$,
On substituting both values to RHS, we get
$$ \Rightarrow RHS = \dfrac{{2{{\sin }^2}\left( {\dfrac{v}{2}} \right)}}{{2\sin \left( {\dfrac{v}{2}} \right)\cos \left( {\dfrac{v}{2}} \right)}}$$,
On further simplifying the above equation we get,
$ \Rightarrow RHS = \dfrac{{\sin \left( {\dfrac{v}{2}} \right)}}{{\cos \left( {\dfrac{v}{2}} \right)}}$,
$ \Rightarrow $$RHS = \tan \left( {\dfrac{v}{2}} \right)$
Therefore, $RHS = LHS$.

Note: There are many ways of solving this problem. The other way of solving this problem is to find the values of $du,dv$ with respect to $x$ and dividing the resulting $du,dv$ values to obtain the answer. So finally, the required answer for the given question is $\dfrac{1}{2}$.