Chapter 14.5, Problem 55PPS

To determine

To Solve :The given value is $\cos \theta \tan \theta -2{\cos }^2\theta =-1$

Answer to Problem 55PPS

  $\theta =(4n-1)\frac{\pi }{2}or\theta =n\pi +{(-1)}^n\frac{\pi }{6},n\in z$

Explanation of Solution

Given information :

We know that, the given expression is $\cos \theta \tan \theta -2{\cos }^2\theta =-1$

Formula Used:

  ${\cos }^2\theta =1-{\sin }^2\theta and\tan \theta =\frac{\sin \theta }{\cos \theta }$

Calculation :

The given equation is

  $\begin{array}{l}\cos \theta \tan \theta -2{\cos }^2\theta =-1\\ \cos \theta \times \frac{\sin \theta }{\cos \theta }-2(1-{\sin }^2\theta )=-1\\ \sin \theta -2+2{\sin }^2\theta =-1\\ 2{\sin }^2\theta +\sin \theta -1=0\\ 2{\sin }^2\theta +2\sin \theta -\sin \theta -1=0\\ (\sin \theta +1)=0or2\sin \theta -1=0\\ \sin \theta =-1or\sin \theta =\frac{1}{2}\\ \theta =(4n-1)\frac{\pi }{2}or\theta =n\pi +{(}^{-}\frac{\pi }{6},n\in z\end{array}$