Prove that \(\displaystyle{\frac{{{\sec{{8}}}{A}-{1}}}{{{\sec{{4}}}{A}-{1}}}}={\frac{{{\tan{{8}}}{A}}}{{{\tan{{2}}}{A}}}}\) I tried converting \(\displaystyle{\sec{{4}}}{A}\)

Sanai Huerta

Sanai Huerta

Answered question

2022-04-01

Prove that sec8A1sec4A1=tan8Atan2A
I tried converting sec4A into 1cos4A and sec8A into 1cos8A but couldn't get the answer. Any help would be appreciated.

Answer & Explanation

Jaslyn Allison

Jaslyn Allison

Beginner2022-04-02Added 13 answers

Using cos2x=12sin2x,sin2x=2sinxcosx
sec2x1tan2x=1cos2xsin2x=2sin2x2sinxcosx=tanx
sec2x1=tan2xtanx
Set 2x=8A,4A and divide

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