You know that the gradient of the tangent at P is , which is slightly more conveniently written as (after multiplying top and bottom by cos(t)). Similarly, the gradients of SP and S'P are . Therefore, by the formula (1), the tan of the angle between the tangent at P and the line SP is
That compound fraction simplifies (I'll skip the details) to
Now use (2) to replace in the denominator by . You will then find that the fraction simplifies further, to
That gives the tan of the angle between the tangent at P and the line SP. What about the analogous calculation for the other focus? Fortunately, you don't need to go through another long computation, because the only difference between the gradient of SP and the gradient of S'P is that e is replaced by -e. So, without any further work, you know that the tan of the angle between the tangent at P and the line S'P is In other words, it is the negative of the previous result. Therefore the two angles are equal (but with opposite sign).