Answer
$$f(x)=\frac{\tan x-1}{\sec x}$$
(a) By using the Quotient Rule to differentiate the function $f$ we have :
$$
\begin{aligned}
f^{\prime}(x)&=\frac{\sec x\left(\sec ^{2} x\right)-(\tan x-1)(\sec x \tan x)}{(\sec x)^{2}}\\
&=\frac{\sec x\left(\sec ^{2} x-\tan ^{2} x+\tan x\right)}{\sec ^{2} x}\\
&=\frac{1+\tan x}{\sec x}
\end{aligned}
$$
(b)
Simplify the expression for $f(x)$ by writing it in terms of $\sin x$ and $\cos x,$ and then find $f^{\prime}(x)$
$$
f(x)=\frac{\tan x-1}{\sec x}=\frac{\frac{\sin x}{\cos x}-1}{\frac{1}{\cos x}}=\frac{\frac{\sin x-\cos x}{\cos x}}{\frac{1}{\cos x}}=\sin x-\cos x
$$
$\Rightarrow$
$$
f^{\prime}(x)=\cos x-(-\sin x)=\cos x+\sin x$$
$$\text { (c) From part (a), } f^{\prime}(x)=\frac{1+\tan x}{\sec x}=\frac{1}{\sec x}+\frac{\tan x}{\sec x}=\cos x+\sin x, \text { which is the expression for } f^{\prime}(x) \text { in part (b) }$$
So, we find that , the expressions for $ f^{\prime}(x)$in parts (a) and (b) are equivalent.
Work Step by Step
$$f(x)=\frac{\tan x-1}{\sec x}$$
(a) By using the Quotient Rule to differentiate the function $f$ we have :
$$
\begin{aligned}
f^{\prime}(x)&=\frac{\sec x\left(\sec ^{2} x\right)-(\tan x-1)(\sec x \tan x)}{(\sec x)^{2}}\\
&=\frac{\sec x\left(\sec ^{2} x-\tan ^{2} x+\tan x\right)}{\sec ^{2} x}\\
&=\frac{1+\tan x}{\sec x}
\end{aligned}
$$
(b)
Simplify the expression for $f(x)$ by writing it in terms of $\sin x$ and $\cos x,$ and then find $f^{\prime}(x)$
$$
f(x)=\frac{\tan x-1}{\sec x}=\frac{\frac{\sin x}{\cos x}-1}{\frac{1}{\cos x}}=\frac{\frac{\sin x-\cos x}{\cos x}}{\frac{1}{\cos x}}=\sin x-\cos x
$$
$\Rightarrow$
$$
f^{\prime}(x)=\cos x-(-\sin x)=\cos x+\sin x$$
$$\text { (c) From part (a), } f^{\prime}(x)=\frac{1+\tan x}{\sec x}=\frac{1}{\sec x}+\frac{\tan x}{\sec x}=\cos x+\sin x, \text { which is the expression for } f^{\prime}(x) \text { in part (b) }$$
So, the expressions for $ f^{\prime}(x)$in parts (a) and (b) are equivalent.