Find the equation of the tangent and normal to the ellipse x²/a²+y²/b²=1 at the point (a cost + b sint?

If you meant, "...at the point (a cost, b sint)", then the answer is as follows:-

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\left(\frac{dy}{dx}\right)=0$

$\Rightarrow \frac{dy}{dx}=-\frac{b^2}{a^2}\left({\displaystyle \frac{x}{y}}\right)$

So, at the point (a cos t, b sin t), ${\overline{)\frac{dy}{dx}}}_{(a \cos t, b \sin t)}=-\frac{b^2}{a^2}\left(\frac{a \cos t}{b \sin t}\right)$

$\Rightarrow {\overline{)\frac{dy}{dx}}}_{(a \cos t, b \sin t)}=-\frac{b}{a}cot t$

Thus, equation of the tangent at (a cos t, b sin t) is:-

$y-b \sin t=-\frac{b}{a}cot t\left(x-a \cos t\right)$

$$\begin{gathered} \Rightarrow ay-ab \sin t + bx cot t -ab \cos t=0 \\ \Rightarrow \left(b cot t\right)x + ay=ab\left(\sin t + \cos t\right) \end{gathered}$$

And, equation of the normal at ​(a cos t, b sin t) is:-

$y-b \sin t=\frac{a}{b} \tan t\left(x-a \cos t\right)$

$$\begin{gathered} \Rightarrow by-b^2\sin t=\left(a \tan t\right)x-a^2\cos t \\ \Rightarrow \left(a \tan t\right)x-by=a^2\cos t-b^2\sin t \end{gathered}$$