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Prove that $\left(1+\tan\left(x\right)\right)\cos\left(x\right)=\sec\left(x\right)^2$ is not an identity

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Final answer to the problem

The equation is not an identity

Step-by-step Solution

How should I solve this problem?

  • Verify if true (using algebra)
  • Express in terms of sine and cosine
  • Simplify
  • Simplify into a single function
  • Express in terms of Sine
  • Express in terms of Cosine
  • Express in terms of Tangent
  • Express in terms of Cotangent
  • Express in terms of Secant
  • Express in terms of Cosecant
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1

Multiplying polynomials $\cos\left(x\right)$ and $1+\tan\left(x\right)$

$\cos\left(x\right)+\cos\left(x\right)\tan\left(x\right)=\sec\left(x\right)^2$

Learn how to solve differential calculus problems step by step online.

$\cos\left(x\right)+\cos\left(x\right)\tan\left(x\right)=\sec\left(x\right)^2$

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Learn how to solve differential calculus problems step by step online. Prove that (1+tan(x))cos(x)=sec(x)^2 is not an identity. Multiplying polynomials \cos\left(x\right) and 1+\tan\left(x\right). Applying the trigonometric identity: \tan\left(\theta \right)\cos\left(\theta \right) = \sin\left(\theta \right). Group the terms of the equation by moving the terms that have the variable x to the left side, and those that do not have it to the right side. Applying the secant identity: \displaystyle\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}.

Final answer to the problem

The equation is not an identity

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Function Plot

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Main Topic: Differential Calculus

The derivative of a function of a real variable measures the sensitivity to change of a quantity (a function value or dependent variable) which is determined by another quantity (the independent variable). Derivatives are a fundamental tool of calculus.

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