Chapter 7.5, Problem 105AYU

a.

To determine

Show that $\tan \left({\tan }^{-1}+{\tan }^{-1}2+{\tan }^{-1}3\right)=0$ .

Answer to Problem 105AYU

  $\boxed{\tan \left[\left({\tan }^{-1}1+{\tan }^{-1}2\right)+{\tan }^{-1}3\right]=0}$

Explanation of Solution

Given information:

Show that $\tan \left({\tan }^{-1}+{\tan }^{-1}2+{\tan }^{-1}3\right)=0$ .

Calculation:

Consider the given expression.

  $\tan \left({\tan }^{-1}+{\tan }^{-1}2+{\tan }^{-1}3\right)=0$

Now group the terms in the parentheses in to two,

  $\tan \left[\left({\tan }^{-1}+{\tan }^{-1}2\right)+{\tan }^{-1}3\right]=0$

Now apply sum formula of tangent.

  $\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

  $\tan \left[\left({\tan }^{-1}1+{\tan }^{-1}2\right)+{\tan }^{-1}3\right]=\frac{\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)+\tan \left({\tan }^{-1}3\right)}{1-\tan \left[\left({\tan }^{-1}1+{\tan }^{-1}2\right)\cdot {\tan }^{-1}3\right]}$

Substitute $3$ for $\tan \left({\tan }^{-1}3\right)$ .

  $\tan \left[\left({\tan }^{-1}1+{\tan }^{-1}2\right)+{\tan }^{-1}3\right]=\frac{\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)+3}{1-\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)\cdot 3}$

Now apply the sum formula of tangent inside the parentheses.

  $\tan \left[\left({\tan }^{-1}1+{\tan }^{-1}2\right)+{\tan }^{-1}3\right]=\frac{\frac{\tan \left({\tan }^{-1}1\right)+\tan \left({\tan }^{-1}2\right)}{1-\tan \left({\tan }^{-1}1\right)\cdot \tan \left({\tan }^{-1}2\right)}+3}{1-\frac{\tan \left({\tan }^{-1}1\right)+\tan \left({\tan }^{-1}2\right)}{1-\tan \left({\tan }^{-1}1\right)\cdot \tan \left({\tan }^{-1}2\right)}\cdot 3}$

Now substitute $1$ for $\tan \left({\tan }^{-1}1\right)$ and $2$ for $\tan \left({\tan }^{-1}2\right)$ we get.

  $\begin{array}{l}\tan \left[\left({\tan }^{-1}1+{\tan }^{-1}2\right)+{\tan }^{-1}3\right]=\frac{\frac{1+2}{1-1\times 2}+3}{1-\frac{1+2}{1-1\times 2}\times 3}\\ =\frac{\frac{3}{-1}+3}{1-\frac{3}{-1}\times 3}=\frac{-3+3}{1+9}\\ \tan \left[\left({\tan }^{-1}1+{\tan }^{-1}2\right)+{\tan }^{-1}3\right]=0\end{array}$

Hence, $\boxed{\tan \left[\left({\tan }^{-1}1+{\tan }^{-1}2\right)+{\tan }^{-1}3\right]=0}$ proved.

b.

To determine

Show that $\tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=\pi$ .

Answer to Problem 105AYU

  $\boxed{{\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3=\pi }$

Explanation of Solution

Given information:

Show that $\tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=\pi$ .

Calculation:

Consider the given expression.

  $\tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=\pi$

It is proved that,

  $\tan \left[\left({\tan }^{-1}+{\tan }^{-1}2\right)+{\tan }^{-1}3\right]=0$

From the definition of inverse tangent function we know that, ${\tan }^{-1}1,{\tan }^{-1}2$ and ${\tan }^{-1}3$ lies in interval $\left(0,\frac{\pi }{2}\right)$

  $\begin{array}{l}0<{\tan }^{-1}1<\frac{\pi }{2}\\ 0<{\tan }^{-1}2<\frac{\pi }{2}\\ 0<{\tan }^{-1}3<\frac{\pi }{2}\end{array}$

So, we have $0<{\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3<3\frac{\pi }{2}$

In the interval ,

  $\begin{array}{l}\left(0,3\frac{\pi }{2}\right),\tan \theta =0if\theta =\pi \\ {\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3=\pi \end{array}$

Hence, $\boxed{{\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3=\pi }$ proved.