Question

Precalculus

Answer 1

\[\frac{ 1 }{ \sec \theta + 1 } + \frac{ 1 }{ \sec \theta - 1 }\]

@Hero @Vocaloid

Simplifying

Answer 2

Hint replace \(\sec\theta\) with \(x\), then add fractions. Replace back with \(\sec \theta\) after simplifying the fraction you added.

Answer 3

By the way, if you didn't get the book Advanced Mathematics with Precalculus by Richard G. Brown, then you're probably studying calculus all wrong. Oh and by the way, you should be able to get through the book in days not years.

Answer 4

\[\frac{ 1 }{ x + 1 } + \frac{ 1 }{ x - 1 }\]
\[\frac{x - 1 }{ (x + 1)(x - 1) } + \frac{ x + 1 }{ (x -1)(x + 1) }\]
\[\frac{ 2x }{ (x + 1)(x - 1) }\]
\[\frac{ 2x }{ x^2 - 1 }\]
\[\frac{ 2\sec \theta }{ \sec^2 \theta - 1 }\]
\[\frac{ \frac{ 2 }{ \cos \theta } }{ \sec^2 \theta - 1 }\]

Since
\[\sec^2 \theta = \tan^2 \theta + 1\]
\[\frac{ \frac{ 2 }{ \cos \theta } }{ (\tan^2 \theta + 1) - 1 }\]
\[\frac{ \frac{ 2 }{ \cos \theta } }{ \tan^2 \theta }\]
\[\frac{ \frac{ 2 }{ \cos \theta } }{ \frac{ \sin^2 \theta }{ \cos^2 \theta } }\]
\[\frac{ 2 }{ \cos \theta } \times \frac{ \cos^2 \theta }{ \sin^2 \theta }\]
\[\frac{ 2 \cos \theta }{ (\sin \theta)(\sin \theta)}\]
\[2\cot \theta \csc \theta\]

Answer 5

Your work looks good.