Trigonometric Identity for tan? Can somebody please help with this (probably simple) query.
Given 
$$dx = R\,d(\tan\theta)$$
this can be expressed as
$$dx = R\,\sec^2\theta\,d\theta$$  
I can't determine where the $\sec^2\theta\,d\theta$ terms are coming from.  I know there is the following Trigonometric Identity for $\tan\theta$ :
$$\tan^2\theta = \sec^2\theta - 1$$
But for some reason I'm struggling!
Not necessarily after the answer but some pointers would be appreciated.
Thankyou.
 A: Recall that the derivative of $\tan\theta$ with respect to $\theta$ is $\sec^2 \theta$. I.e., $$\dfrac{d}{d\theta}\left(\tan \theta\right) = \sec^2 \theta$$
So we have $$d(\tan\theta) = \sec^2 \theta\,d\theta$$
A: compute the derivative of the ratio $\tan = \sin/\cos$:
$$
\tan' = \frac{\sin'\cos - \cos'\sin}{\cos^2}
= \frac{\cos^2 + \sin^2}{\cos^2}=\frac1{\cos^2} = \sec^2
\\
dx = R\,d(\tan\theta) = R\tan'\theta \,d\theta = R\sec^2\theta \,d\theta
$$
A: An argument currently in the queue for less-than-one-page notes in the American Mathematical Monthly goes like this (but you'll have to draw the pictures yourself or wait for the publication):
$$
\tan\theta = \frac{\text{opposite}}{\text{adjacent}}= \frac{\text{opposite}}{1}, \qquad \sec\theta=\frac{\text{hypotenuse}}{\text{adajcent}}= \frac{\text{hypotenuse}}{1} \tag 1
$$
Let $\theta$ change by an infinitely small amount $d\theta$ while the "adjacent" side remains equal to $1$.  This results in an infinitely small change $d\tan\theta$ in $\tan\theta$.  The infinitely small segment from height $\tan\theta$ to height $\tan\theta+d\tan\theta$ on the "opposite" side is one side of a small triangle.  A segment going from the point at height $\tan\theta+d\tan\theta$ on the "opposite" side in the direction toward the vertex of the angle whose measure is $\theta+d\theta$ is another side of the small triangle.  The third side is an arc of a circle centered at the aforementioned vertex, that passes through the point at height $\tan\theta$ on the "opposite" side.  Since it is an infinitely small arc, it is a straight line.  The length of that arc is $r\,d\theta=\sec\theta\,d\theta$.  This small triangle has the same three angles as the big triangle whose sides are referred to in $(1)$ above; hence the same ratios of lengths of sides:
$$
\frac{d\tan\theta}{\sec\theta\,d\theta} = \frac{\sec\theta}{1}.
$$
We conclude that $\dfrac d{d\theta}\tan\theta=\sec^2\theta$. (And we get the derivative of the secant function from the another pair of sides of the small triangle.)
So that's one way of seeing why $d\tan\theta$ must be $\sec^2\theta\,d\theta$.
