(a) use the quotient rule to differentiate the function\nf(x)=\frac{\tan(x)-1}{sec(x)}.\nf(x)=\n

Explanation:

Step1: Recall the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. For $f(x)=\frac{\tan(x)-1}{\sec(x)}$, let $u = \tan(x)-1$ and $v=\sec(x)$.

Step2: Find $u'$ and $v'$

We know that the derivative of $\tan(x)$ is $\sec^{2}(x)$ and the derivative of a constant (in this case, $- 1$) is $0$, so $u'=\sec^{2}(x)$. The derivative of $\sec(x)$ is $\sec(x)\tan(x)$, so $v'=\sec(x)\tan(x)$.

Step3: Apply the quotient - rule

[ \begin{align*} f'(x)&=\frac{\sec^{2}(x)\cdot\sec(x)-(\tan(x) - 1)\cdot\sec(x)\tan(x)}{\sec^{2}(x)}\ &=\frac{\sec^{3}(x)-\sec(x)\tan^{2}(x)+\sec(x)\tan(x)}{\sec^{2}(x)}\ &=\frac{\sec(x)(\sec^{2}(x)-\tan^{2}(x)+\tan(x))}{\sec^{2}(x)}\ \end{align*} ] Since $\sec^{2}(x)-\tan^{2}(x) = 1$, we have: [ \begin{align*} f'(x)&=\frac{\sec(x)(1 + \tan(x))}{\sec^{2}(x)}\ &=\frac{1+\tan(x)}{\sec(x)}\ &=\cos(x)(1 + \tan(x))\ &=\cos(x)+\sin(x) \end{align*} ]

Answer:

$\cos(x)+\sin(x)$