Expert Answer to "Prove the trigonometry identitied sec⁡Asec⁡A−1−sec⁡A+1tan2A=1"

Harlen Pritchard

Answered question

2021-09-10

Prove the trigonometry identitied

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

faldduE

Skilled2021-09-11Added 109 answers

Solve the LHS. Use the identity

(a + b) (a - b) = a^{2} - b^{2}

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

= \frac{\sec A \tan^{2} A - (\sec A + 1) (\sec A - 1)}{(\sec A - 1) \tan^{2} A}

= \frac{\sec A \tan^{2} A - (\sec^{2} A - 1)}{(\sec A - 1) \tan^{2} A}

Use the identity

\sec^{2} θ - 1 = \tan^{2} θ

LHS = \frac{\sec A \tan^{2} A - \tan^{2} A}{(\sec A - 1) \tan^{2} A}

= \frac{\tan^{2} A (\sec A - 1)}{(\sec A - 1) \tan^{2} A}

= 1

= RHS

Since LHS=RHS. Hence

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

proved.

Jeffrey Jordon

Expert2022-01-31Added 2605 answers

Answer is given below (on video)

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