Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 660: 66


The solution of the given question is $2\csc x$.

Work Step by Step

$\frac{1+\cos x}{\sin x}+\frac{\sin x}{1+\cos x}$ Then, use a common denominator for the given equation: $\begin{align} & \frac{1+\cos x}{\sin x}+\frac{\sin x}{1+\cos x}=\frac{1+\cos }{\sin x}\times \frac{1+\cos x}{1+\cos x}+\frac{\sin x}{1+\cos x}\times \frac{\sin x}{\sin x} \\ & =\frac{1+2\cos x+{{\cos }^{2}}x}{\left( \sin x \right)\left( 1+\cos x \right)}+\frac{{{\sin }^{2}}x}{\left( \sin x \right)\left( 1+\cos x \right)} \end{align}$ The two fractions will be simplified by creating a common base as shown below: $\begin{align} & \frac{1+2\cos x+{{\cos }^{2}}x}{\left( \sin x \right)\left( 1+\cos x \right)}+\frac{{{\sin }^{2}}x}{\left( \sin x \right)\left( 1+\cos x \right)}=\frac{1+2\cos x+{{\cos }^{2}}x+{{\sin }^{2}}x}{\left( \sin x \right)\left( 1+\cos x \right)} \\ & =\frac{1+2\cos x+1}{\left( \sin x \right)\left( 1+\cos x \right)} \\ & =\frac{2+2\cos x}{\left( \sin x \right)\left( 1+\cos x \right)} \\ & =\frac{2\left( 1+\cos x \right)}{\left( \sin x \right)\left( 1+\cos x \right)} \end{align}$ By using the reciprocal identity $\frac{1}{\sin x}=\csc x$, we get: $\begin{align} & \frac{2\left( 1+\cos x \right)}{\left( \sin x \right)\left( 1+\cos x \right)}=\frac{2}{\sin x} \\ & =\frac{2}{\sin x} \\ & =2\csc x \end{align}$ Conjecture: Left side is equal to $2\csc x$. Thus, the left side of the expression is equal to the right side, which is $\frac{1+\cos x}{\sin x}+\frac{\sin x}{1+\cos x}=2\csc x$.
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