I don't understand line 3.
After using the Quotient Rule, how do I use the Trig identities to simplify the third line?
Also, what is the strategy when using trig identities, when do you know you are finished?
If I can choose between $\cot \ x = \frac{\cos \ x}{\sin \ x}$, which one would I choose? $\cot \ x$, or $\frac{\cos \ x}{\sin \ x}$?
Differentiate $f(x) = \dfrac{\sec x}{1+\tan x}.\;$For what values of $x$ does the graph of $f$ have a horizontal tangent?
Solution The Quotient Rule gives $$\begin{align} f'(x) & = \frac{(1+\tan x)\frac d{dx}(\sec x) - \sec x \frac d{dx}(1+\tan x)}{(1+\tan x)^2}\\ \\ & = \frac{(1+\tan x)\sec x \tan x - \sec x \cdot \sec^2 x}{(1+\tan x)^2}\\\\ &= \frac{\sec x(\tan x + \tan^2 x - \sec^2 x)}{(1+\tan s)^2}\\ \\ &= \frac{\sec x(\tan x - 1)}{(1 + \tan x)^2} \end{align}$$
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$\begingroup$Question 1: Third Line Issue $\tan ^2x - \sec ^2 x = \left(\frac{\sin x}{\cos x}\right)^2-\left(\frac{1}{\cos x}\right)^2=\frac{\sin ^2x-1}{\cos ^2x}=\frac{-\cos ^2x}{\cos ^2x}=-1$
Question 2: General Question It is a matter of choice really. Inasmuch as you are talking about equivalent expressions, then these expression are equally valid. That said, one often chooses the form that is more "concise," whatever "concise" means to you.
So, in the case you mentioned, $\cot x$ versus $\frac{\cos x}{\sin x}$, which seems more concise to you? Moreover, the final expression is equivalent to $\frac{\sin x-\cos x}{\left(\sin x+\cos x\right)^2}$. Does that appear more elegant than Line 4 of the post?
$\endgroup$ 3 $\begingroup$To answer one of your querry:
Factor out $sec\ x$ and multiply $(1+tan\ x)tan\ x$ in the numerator of line 2. We now have the third line. Using the identity $tan^2+1=sec^2x$ we have line 4.
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