Chapter 14.5, Problem 23CYU

To determine

To find :The value of $\theta$ .

Answer to Problem 23CYU

  $\theta =\frac{3\pi }{4}$

Explanation of Solution

Given information :

We know that, the given information is ${\tan }^2\theta +\theta \tan \theta +1=0$

Formula Used :

  $a^2+b^2+2ab={(a+b)}^2$

Calculation :

The given equation is

  $\begin{array}{l}{\tan }^2\theta +2\tan \theta +1=0\\ $ \Rightarrow {(\tan \theta +1)}^2=0$\Rightarrow (\tan \theta +1)(\tan \theta +1)=0\\ \Rightarrow \tan \theta +1=0or\tan \theta +1=0\\ \Rightarrow \tan \theta =-1\\ \Rightarrow \tan \theta =-\tan \frac{\pi }{4}\\ \Rightarrow \tan \theta =\tan \left(\pi -\frac{\pi }{4}\right)\\ \Rightarrow \tan \theta =\tan \frac{3\pi }{4}\\ \Rightarrow \theta =\frac{3\pi }{4}\end{array}$