Find the Exact Value cos(arctan(x))

$\cos (\arctan (x))$

Draw a triangle in the plane with vertices $(1, x)$ , $(1, 0)$ , and the origin. Then $\arctan (x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, x)$ . Therefore, $\cos (\arctan (x))$ is $\frac{1}{\sqrt{1 + x^{2}}}$ .

$\frac{1}{\sqrt{1 + x^{2}}}$

Multiply $\frac{1}{\sqrt{1 + x^{2}}}$ by $\frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$ .

$\frac{1}{\sqrt{1 + x^{2}}} \cdot \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{1 + x^{2}}}$ by $\frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$ .

$\frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}} \sqrt{1 + x^{2}}}$

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$\frac{\sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1} \sqrt{1 + x^{2}}}$

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$\frac{\sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1} {\sqrt{1 + x^{2}}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1 + 1}}$

Add $1$ and $1$ .

$\frac{\sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{2}}$

Rewrite ${\sqrt{1 + x^{2}}}^{2}$ as $1 + x^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 + x^{2}}$ as ${(1 + x^{2})}^{\frac{1}{2}}$ .

$\frac{\sqrt{1 + x^{2}}}{{({(1 + x^{2})}^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{\sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\frac{\sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{\sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}}$

Rewrite the expression.

$\frac{\sqrt{1 + x^{2}}}{{(1 + x^{2})}^{1}}$

Simplify.

$\frac{\sqrt{1 + x^{2}}}{1 + x^{2}}$

Find the Exact Value cos(arctan(x))

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