find an equation of the tangent line to the curve at the given point.\n\n$y = \\frac{1 + \\sin(x)}{\\cos(x)}$…

Explanation:

Step1: Differentiate ( y=\frac{1 + \sin(x)}{\cos(x)} ) using the quotient rule

The quotient rule is ( \left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}} ). Let ( u = 1+\sin(x) ), ( u'=\cos(x) ); ( v=\cos(x) ), ( v'=-\sin(x) ). Then ( y'=\frac{\cos(x)\cdot\cos(x)-(1 + \sin(x))\cdot(-\sin(x))}{\cos^{2}(x)}=\frac{\cos^{2}(x)+\sin(x)+\sin^{2}(x)}{\cos^{2}(x)} ). Using ( \sin^{2}(x)+\cos^{2}(x) = 1 ), we get ( y'=\frac{1+\sin(x)}{\cos^{2}(x)} ).

Step2: Evaluate the derivative at ( x = \pi )

Substitute ( x=\pi ) into ( y' ). ( y'(\pi)=\frac{1+\sin(\pi)}{\cos^{2}(\pi)} ). Since ( \sin(\pi)=0 ) and ( \cos(\pi)=- 1 ), then ( y'(\pi)=\frac{1 + 0}{(-1)^{2}}=1 ).

Step3: Use the point - slope form ( y - y_{1}=m(x - x_{1}) )

Here ( m = 1 ), ( x_{1}=\pi ), ( y_{1}=-1 ). So ( y+1=1\cdot(x - \pi) ). Simplifying gives ( y=x-\pi - 1 ).

Answer:

( y=x-\pi - 1 )