Chapter 7.2, Problem 20E

To determine

To verify: that the given equation is an identity.

Answer to Problem 20E

See explanation

Explanation of Solution

Given information:

  $\left(\sin \theta -1\right)\left(\tan \theta +\sec \theta \right)=-\cos \theta$

Calculation:

Simplify the left side of the equation into right side as follows:

  $\begin{array}{c}\left(\sin \theta -1\right)\left(\tan \theta +\sec \theta \right)=-\cos \theta \end{array}$

  $\begin{array}{c}\left(\sin \theta -1\right)\left(\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }\right)=-\cos \theta \left[\begin{array}{l}\text{Quotient identity: }\tan x=\frac{\sin x}{\cos x}\\ \text{Reciprocal identity: }\sec x=\frac{1}{\cos x}\end{array}\right]\end{array}$

  $\begin{array}{c}\frac{{\sin }^2\theta }{\cos \theta }+\frac{\sin \theta }{\cos \theta }-\frac{\sin \theta }{\cos \theta }-\frac{1}{\cos \theta }=-\cos \theta \end{array}$

  $\begin{array}{c}\frac{{\sin }^2\theta }{\cos \theta }-\frac{1}{\cos \theta }=-\cos \theta \end{array}$

  $\begin{array}{c}\frac{{\sin }^2\theta -1}{{\cos }^2\theta }=-\cos \theta \end{array}$

  $\begin{array}{c}\frac{-{\cos }^2\theta }{\cos \theta }=-\cos \theta \left[{\text{Pythagorean identity: sin}}^{2}x+{\cos }^{2}x=1\right]\end{array}$

  $\begin{array}{c}-\cos \theta =-\cos \theta \end{array}$

  $\begin{array}{c}LHS=RHS\end{array}$

Since the left side of the equation is transformed into right hand side. So, the equation is an identity.