Editing Squeeze theorem (section)

=== Third example ===

It is possible to show that
<math display="block"> \frac{d}{d\theta} \tan\theta = \sec^2\theta </math>
by squeezing, as follows.

[[File:Tangent.squeeze.svg|thumb|upright=1.5|right]]

In the illustration at right, the area of the smaller of the two shaded sectors of the circle is

<math display="block"> \frac{\sec^2\theta\,\Delta\theta}{2}, </math>

since the radius is {{math|sec ''θ''}} and the arc on the [[unit circle]] has length&nbsp;{{math|Δ''θ''}}.  Similarly, the area of the larger of the two shaded sectors is

<math display="block"> \frac{\sec^2(\theta + \Delta\theta)\,\Delta\theta}{2}. </math>

What is squeezed between them is the triangle whose base is the vertical segment whose endpoints are the two dots. The length of the base of the triangle is {{math|tan(''θ'' + Δ''θ'') − tan ''θ''}}, and the height is&nbsp;1. The area of the triangle is therefore

<math display="block"> \frac{\tan(\theta + \Delta\theta) - \tan\theta}{2}. </math>

From the inequalities

<math display="block"> \frac{\sec^2\theta\,\Delta\theta}{2} \le \frac{\tan(\theta + \Delta\theta) - \tan\theta}{2} \le \frac{\sec^2(\theta + \Delta\theta)\,\Delta\theta}{2} </math>

we deduce that

<math display="block"> \sec^2\theta \le \frac{\tan(\theta + \Delta\theta) - \tan\theta}{\Delta\theta} \le \sec^2(\theta + \Delta\theta),</math>

provided&nbsp;{{math|Δ''θ'' > 0}}, and the inequalities are reversed if&nbsp;{{math|Δ''θ'' < 0}}.  Since the first and third expressions approach {{math|sec<sup>2</sup>''θ''}} as {{math|Δ''θ'' → 0}}, and the middle expression approaches <math>\tfrac{d}{d\theta} \tan\theta,</math> the desired result follows.