Chapter 6, Problem 13CLT

To determine

Subtract and simplify $\frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}$ .

Answer to Problem 13CLT

Simplified expression to $\frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}\text{is }2\tan \theta$

Explanation of Solution

Given info:

  $\frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}$

Formula used:

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

We need to find $\frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}$

Calculating

  $\begin{array}{l}\frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}=\frac{\left(\sin \theta -1\right)\left(\sin \theta -1\right)-{\cos }^2\theta }{\cos \theta \left(\sin \theta -1\right)}\\ \frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}=\frac{{\sin }^2\theta -2\sin \theta +1-{\cos }^2\theta }{\cos \theta \left(\sin \theta -1\right)}\\ \frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}=\frac{2{\sin }^2\theta -2\sin \theta }{\cos \theta \left(\sin \theta -1\right)}\\ \frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}=\frac{2\sin \theta \left(\sin \theta -1\right)}{\cos \theta \left(\sin \theta -1\right)}\\ \frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}=\frac{2\sin \theta }{\cos \theta }\\ \frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}=2\tan \theta \end{array}$

Simplified expression to $\frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}\text{is }2\tan \theta$

Conclusion:

Thus,simplified expression to $\frac{\sin \theta -1}{\cos \theta }-\frac{\cos \theta }{\sin \theta -1}\text{is }2\tan \theta$