Why \(\displaystyle{\frac{{{{\sec}^{{2}}{x}}}}{{\sqrt{{{{\sec}^{{2}}{x}}}}}}}\ne{\sec{{x}}}\)? When I evaluate \(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({{\text{sinh}}^{{-{1}}}{\left({\tan{{x}}}\right)}}\right)}\), i

Paula Good

Paula Good

Answered question

2022-03-22

Why sec2xsec2xsecx?
When I evaluate ddx(sinh1(tanx)), i get:
sec2x1+tan2x
At this point I want to use the sec2x=1+tan2x identity to simplify to:
sec2xsecx=secx

Answer & Explanation

haiguetenteme7zyu

haiguetenteme7zyu

Beginner2022-03-23Added 13 answers

Formally, you get
sec2x1+tan2x=sec2xsec2x=sec2x|secx|=|secx|

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?