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The equation of the tangent to the unit circle at point ($\cos \alpha, \sin \alpha $) is

  1. $x\cos \alpha-y \sin\alpha=1 $
  2. $x\sin \alpha-y \cos\alpha =1$
  3. $x\cos \alpha+ y\sin\alpha=1 $
  4. $x\sin \alpha-y \cos\alpha=1 $
  5. None of the above
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Assuming that the unit circle is centered at the origin, the equation of the unit circle is: $x^2 + y^2 = 1$

The slope of the tangent to the unit circle at point $(x,y)$ can be derived by implicit differentiation as follows:

$$\begin{align}
x^2 + y^2 &= 1 \\[1em]
\frac{d}{dx} \left ( x^2 + y^2 \right ) &= \frac{d}{dx}1 \\[1em]
2x + 2y \frac{dy}{dx} &= 0 \\[1em]
\frac{dy}{dx} &= -\frac xy
\end{align}$$

Thus, the slope of the tangent at the point $(\cos \alpha, \sin \alpha)$ is $-\dfrac{\cos \alpha}{\sin \alpha}$

The equation of the tangent line then will be:

$$\begin{align}
y - y_1 &= m(x - x_1)\\[1em]
y - \sin \alpha &= -\dfrac{\cos \alpha}{\sin \alpha}(x - \cos \alpha)\\[1em]
y \sin \alpha - \sin^2 \alpha &= -x \cos \alpha + \cos^2 \alpha\\[1em]
y \sin \alpha + x \cos \alpha &= \sin^2 \alpha + \cos^2 \alpha = 1
\end{align}$$

Hence, option c is correct.
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