Answer:
[tex]{ \bf{f(x) = \tan x + 9 \sin x }}[/tex]
For gradient, differentiate f(x):
[tex]{ \tt{ \frac{dy}{dx} = { \sec }^{2}x + 9 \cos x }}[/tex]
Substitute for x as π:
[tex]{ \tt{gradient = { \sec }^{2} \pi + 9 \cos(\pi ) }} \\ { \tt{gradient = - 8 }}[/tex]
Gradient of tangent = -8
[tex]{ \bf{y =mx + b }} \\ { \tt{0 = ( - 8\pi) + b}} \\ { \tt{b = 8\pi}} \\ y - intercept = 8\pi[/tex]
Equation of tangent:
[tex]{ \boxed{ \bf{y = - 8x + 8\pi}}}[/tex]
