How do you simplify $\sqrt{1 + {tan}^{2} x}$ $sqrt(1+tan^2x)$ ?

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How do you simplify $\sqrt{1 + {tan}^{2} x}$ $sqrt(1+tan^2x)$ ?

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Answer 1 · 2017-01-13T20:13:20

$\sqrt{1 + {tan}^{2} x} = | sec x |$ $sqrt(1+tan^2 x) = abs(sec x)$

Explanation:

Using:

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2 x + sin^2 x = 1$

$tan x = \frac{sin x}{cos x}$ $tan x = sin x / cos x$

$sec x = \frac{1}{cos x}$ $sec x = 1/cos x$

we find:

$\sqrt{1 + {tan}^{2} x} = \sqrt{1 + \frac{{sin}^{2} x}{{cos}^{2} x}}$ $sqrt(1+tan^2 x) = sqrt(1+(sin^2 x)/(cos^2 x))$

$\sqrt{1 + {tan}^{2} x} = \sqrt{\frac{{cos}^{2} x}{{cos}^{2} x} + \frac{{sin}^{2} x}{{cos}^{2} x}}$ $color(white)(sqrt(1+tan^2 x)) = sqrt((cos^2 x)/(cos^2 x)+(sin^2 x)/(cos^2 x))$

$\sqrt{1 + {tan}^{2} x} = \sqrt{\frac{{cos}^{2} x + {sin}^{2} x}{{cos}^{2} x}}$ $color(white)(sqrt(1+tan^2 x)) = sqrt((cos^2 x+sin^2 x)/(cos^2 x))$

$\sqrt{1 + {tan}^{2} x} = \sqrt{\frac{1}{{cos}^{2} x}}$ $color(white)(sqrt(1+tan^2 x)) = sqrt(1/(cos^2 x))$

$\sqrt{1 + {tan}^{2} x} = \sqrt{{sec}^{2} x}$ $color(white)(sqrt(1+tan^2 x)) = sqrt(sec^2 x)$

$\sqrt{1 + {tan}^{2} x} = | sec x |$ $color(white)(sqrt(1+tan^2 x)) = abs(sec x)$

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How do you simplify $\sqrt{1 + {tan}^{2} x}$ $sqrt(1+tan^2x)$ ?

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