How to simplify this trigonometric expression?

((sinx+tanx)^2+cos^2x-sec^2x)/tanx
Answers

la console

= { [sin(x) + tan(x)]² + cos²(x) - sec²(x) } / tan(x) → you expand = { sin²(x) + 2.sin(x).tan(x) + tan²(x) + cos²(x) - sec²(x) } / tan(x) → recall: sin²(x) + cos²(x) = 1 = { 1 + 2.sin(x).tan(x) + tan²(x) - sec²(x) } / tan(x) = [1/tan(x)] + [2.sin(x).tan(x)/tan(x)] + [tan²(x)/tan(x)] - [sec²(x)/tan(x)] → you simplify = [1/tan(x)] + 2.sin(x) + tan(x) - [sec²(x)/tan(x)] → recall: sec(x) = 1/cos(x) = [1/tan(x)] + 2.sin(x) + tan(x) - [{1/cos²(x)}/tan(x)] → recall: tan(x) = sin(x)/cos(x) = [1/tan(x)] + 2.sin(x) + tan(x) - [{1/cos²(x)} / {sin(x)/cos(x)}] = [1/tan(x)] + 2.sin(x) + tan(x) - [{1/cos²(x)} * {cos(x)/sin(x)}] = [1/tan(x)] + 2.sin(x) + tan(x) - [{1 * cos(x)} / {cos²(x) * sin(x)}] → you simplify = [1/tan(x)] + 2.sin(x) + tan(x) - [1/{cos(x).sin(x)}] → recall: tan(x) = sin(x)/cos(x) = [cos(x)/sin(x)] + 2.sin(x) + [sin(x)/cos(x)] - [1/{cos(x).sin(x)}] → common denominator: cos(x).sin(x) = [cos²(x) + 2.sin²(x).cos(x) + sin(x)² - 1] / [cos(x).sin(x)] → recall: sin²(x) + cos²(x) = 1 = [1 + 2.sin²(x).cos(x) - 1] / [cos(x).sin(x)] → you can simplify = [2.sin²(x).cos(x)] / [cos(x).sin(x)] → you can simplify by cos(x) = [2.sin²(x)] / sin(x) → you can simplify by sin(x) = 2.sin(x)

Captain Matticus, LandPiratesInc

sin(x)^2 + 2sin(x)tan(x) + tan(x)^2 + cos(x)^2 - sec(x)^2 => sin(x)^2 + cos(x)^2 - (sec(x)^2 - tan(x)^2) + 2 * sin(x) * tan(x) => 1 - 1 + 2 * sin(x) * tan(x) => 2 * sin(x) * tan(x) 2 * sin(x) * tan(x) / tan(x) => 2 * sin(x)