Subsection 2.8.6 Step 4: the Remaining Trigonometric Functions
It is now an easy matter to get the derivatives of the remaining trigonometric functions using basic trig identities and the quotient rule. Remember 8 You really should. If you do not then take a quick look at the appropriate appendix. that
\begin{align*}
\tan x&= \frac{\sin x}{\cos x} & \cot x &= \frac{\cos x}{\sin x}=
\frac{1}{\tan x}\\
\csc x&= \frac{1}{\sin x} & \sec x &= \frac{1}{\cos x}
\end{align*}
So, by the quotient rule,
\begin{alignat*}{3}
\diff{}{x}\tan x
&=\diff{}{x}\,\frac{\sin x}{\cos x} &
&=\frac{\overbrace{\big({\ts\diff{}{x}}\sin x\big)}^{\cos x}\cos x
-\sin x\overbrace{\big({\ts\diff{}{x}\cos x}\big)}^{-\sin x}}
{\cos^2 x} &
&=\sec^2 x\\
\diff{}{x}\,\csc x
&=\diff{}{x}\,\frac{1}{\sin x} &
&=-\frac{\overbrace{\big({\ts\diff{}{x}}\sin x\big)}^{\cos x}}
{\sin^2 x} &
&=-\csc x\cot x\\
\diff{}{x}\,\sec x
&=\diff{}{x}\,\frac{1}{\cos x} &
&=-\frac{\overbrace{\big({\ts\diff{}{x}}\cos x\big)}^{-\sin x}}
{\cos^2 x} &
&=\sec x\tan x\\
\diff{}{x}\cot x
&=\diff{}{x}\,\frac{\cos x}{\sin x} &
&=\frac{ \overbrace{\big({\ts\diff{}{x}}\cos x\big)}^{-\sin x}\sin x
-\cos x\overbrace{{\ts\big(\diff{}{x}}\sin x\big)}^{\cos x}}
{\sin^2 x} &
&=-\csc^2 x
\end{alignat*}