Question

In Exercises 67–74, rewrite each expression in terms of the given function or functions. 11−cos⁡x−cos⁡x1+cos⁡x\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}; csc⁡x\csc x

Accepted Answer

Identify the given expression: $$ \frac{\cos x}{1 - \cos x} - \frac{1}{\csc x (1 + \cos x)} . $$Our goal is to rewrite this expression in terms of sine and cosine functions, or in terms of the given trigonometric functions. Recall that $$ \csc x = \frac{1}{\sin x} . $$Substitute this into the expression to rewrite the denominator of the second term: $$ \frac{1}{\csc x (1 + \cos x)} = \frac{1}{\frac{1}{\sin x} (1 + \cos x)} = \frac{1}{\frac{1 + \cos x}{\sin x}} . $$Simplify the second term by taking the reciprocal of the denominator: $$ \frac{1}{\frac{1 + \cos x}{\sin x}} = \frac{\sin x}{1 + \cos x} . $$Now the expression becomes $$ \frac{\cos x}{1 - \cos x} - \frac{\sin x}{1 + \cos x} . To $$combine the two terms, find a common denominator, which is $$ (1 - \cos x)(1 + \cos x) . $$Recall the Pythagorean identity $$ (1 - \cos x)(1 + \cos x) = 1 - \cos^2 x = \sin^2 x . $$Rewrite each term with the common denominator $$ \sin^2 x $$: $$ \frac{\cos x (1 + \cos x)}{\sin^2 x} - \frac{\sin x (1 - \cos x)}{\sin^2 x} . $$From here, you can combine the numerators over the common denominator and simplify further if needed.

In Exercises 67–74, rewrite each expression in terms of the given function or functions. $\frac{1}{1 - \cos x} - \frac{\cos x}{1 + \cos x}$ ; $\csc x$

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Concetti chiave

Reciprocal Trigonometric Functions

Pythagorean Identities

Algebraic Manipulation of Trigonometric Expressions

In Exercises 67–74, rewrite each expression in terms of the given function or functions. 11−cosx−cosx1+cosx\(\frac{1}{1-\cos x}\)-\(\frac{\cos x}{1+\cos x}\); cscx\(\csc\) x