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  "body": "\n## Proof\n\n1. Let $a$ be real with $\\cos a \\ne 0$. By the [[definition of tangent|06T0HS14PP]], $\\tan x = \\frac{\\sin x}{\\cos x}$ wherever $\\cos x \\ne 0$.\n2. By the [[derivative of cosine|0V2DV29EGT]], $\\cos$ is differentiable, hence [[continuous|07ZMX4ANJS]]. So there is an open interval $I$ containing $a$ on which $\\cos x \\ne 0$.\n3. On $I$, $\\tan = \\frac{\\sin}{\\cos}$ is a quotient of two functions that are differentiable at $a$, by the [[derivative of sine|00R4MWW532]] and the derivative of cosine.\n4. By the [[quotient rule|0BS0NJPKVE]], $\\tan$ is differentiable at $a$, and $\\tan'(a) = \\frac{\\cos a \\cdot \\cos a - \\sin a \\cdot (-\\sin a)}{\\cos^2 a} = \\frac{\\cos^2 a + \\sin^2 a}{\\cos^2 a}$.\n5. By the [[Pythagorean identity|03H7WZ4T3K]], the numerator is $1$. So $\\tan'(a) = \\frac{1}{\\cos^2 a}$.\n6. Splitting the fraction in step 4 instead gives $\\tan'(a) = 1 + \\frac{\\sin^2 a}{\\cos^2 a} = 1 + \\tan^2 a$.\n7. So $\\tan$ is differentiable at every $a$ with $\\cos a \\ne 0$, and $\\tan'(a) = \\frac{1}{\\cos^2 a} = 1 + \\tan^2 a$.\n\n## Idea\n\nTangent is sine over cosine, so the quotient rule applies wherever cosine is not zero. The numerator becomes $\\cos^2 + \\sin^2$, which is $1$ by the Pythagorean identity. Dividing it term by term instead gives the form $1 + \\tan^2$.\n",
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