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tan^(-1)((1)/(8x+9))

tan1(18x+9) \tan ^{-1}\left(\frac{1}{8 x+9}\right)

Full solution

Q. tan1(18x+9) \tan ^{-1}\left(\frac{1}{8 x+9}\right)
  1. Understand the problem: Understand the problem.\newlineWe need to simplify the expression tan1(18x+9)\tan^{-1}\left(\frac{1}{8x+9}\right). This is the inverse tangent (also known as arctan) of the fraction 18x+9\frac{1}{8x+9}.
  2. Recognize properties: Recognize the properties of the inverse tangent function.\newlineThe inverse tangent function, tan1(x)\tan^{-1}(x), does not have algebraic simplifications like the ones we use for powers or roots. It is a trigonometric function that gives us an angle whose tangent is xx. Therefore, the expression tan1(18x+9)\tan^{-1}\left(\frac{1}{8x+9}\right) is already in its simplest form as a function of xx.
  3. Check for simplifications: Check for any possible simplifications. Since the expression inside the inverse tangent function is a simple fraction and cannot be simplified further, and there are no trigonometric identities that apply to inverse tangent functions with a variable inside, we conclude that the expression tan1(18x+9)\tan^{-1}\left(\frac{1}{8x+9}\right) is already in its simplest form.

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