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Chapter 29: Problem 957

Prove that \(1 /(\sec A-\tan A)=\sec A+\tan A\) is an identity.

Short Answer

Expert verified
To prove the identity \(1 /(\sec A-\tan A)=\sec A+\tan A\), start by rewriting the given expression in terms of sine and cosine functions: \(1 /(\frac{1}{\cos A}-\frac{\sin A}{\cos A})\). Then, simplify by finding a common denominator and multiplying the expression by the appropriate factors. The result, \(\frac{1 + \sin A}{\cos A}\), is equal to the given expression \(\sec A + \tan A\), proving the identity.

Step by step solution

01

Rewrite the expression in terms of sine and cosine functions

Recall that \(\sec A = \frac{1}{\cos A}\) and \(\tan A = \frac{\sin A}{\cos A}\). Then the given expression \(1 /(\sec A-\tan A)\) can be rewritten as: \(1 /(\frac{1}{\cos A}-\frac{\sin A}{\cos A})\)
02

Simplify the expression

To simplify the expression, find a common denominator for the terms in the parentheses: \(1 /(\frac{1 - \sin A}{\cos A})\) Now, multiply the numerator and denominator of the fraction by \(\cos A\): \((\cos A) / (1 - \sin A)\)
03

Show that the expression is equal to \(\sec A + \tan A\)

To show that the simplified expression is equal to \(\sec A + \tan A\), let's rewrite \(\sec A + \tan A\) in terms of sine and cosine functions: \(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\) Now, combine the terms with the same denominator and simplify: \(\frac{1 + \sin A}{\cos A}\) Notice that if we multiply the numerator and denominator of the simplified expression from step 2 by \(-1\), we get the expression from this step: \((-\cos A) / (-(1 - \sin A)) = \frac{\cos A}{1 + \sin A} = \frac{1 + \sin A}{\cos A}\). Hence, we can conclude that \(1 /(\sec A-\tan A) = \sec A+\tan A\), and thus the given expression is indeed an identity.

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