"Proving 1sec⁡ϕ−tan⁡ϕ−1sec⁡ϕ+tan⁡ϕ=2tan⁡ϕ" solved and explained - Plainmath

lugreget9

Answered question

2022-01-17

Proving

\frac{1}{\sec ϕ - \tan ϕ} - \frac{1}{\sec ϕ + \tan ϕ} = 2 \tan ϕ

Answer & Explanation

Jeffery Autrey

Beginner2022-01-18Added 35 answers

HINT:
Obtain a common denominator to arrive at

\frac{1}{\sec ϕ - \tan ϕ} - \frac{1}{\sec ϕ + \tan ϕ} = \frac{2 \tan ϕ}{\sec^{2} ϕ - \tan^{2} ϕ}

RizerMix

Skilled2022-01-19Added 437 answers

$\frac{1}{\sec ϕ - \tan ϕ} - \frac{1}{\sec ϕ + \tan ϕ} = \frac{\cos ϕ}{1 - \sin ϕ} - \frac{\cos ϕ}{1 + \sin ϕ} = \cos ϕ \underset{\cos^{2} ϕ}{\underset{⏟}{\frac{2 \sin ϕ}{1 - \sin^{2} ϕ}}} = 2 \frac{\sin ϕ}{\cos ϕ}$

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