<s>[INST] Trouble with tedious algebra (Oxford 1992 Admissions Test 2 1992) 
(i) Show that the condition that the points $P$ $(a\cos A,b\sin A)$ and $Q$ $(a\cos B,b\sin  B )$
should subtend a right angle at O is
$$a^2\cos A\cos B+b^2\sin A\sin  B=0$$
(ii) Let S be a circle centre $O$ and radius $C$. Find the equation of the tangent to S at
the point $(C\cos  t, C\sin t)$.
(iii) If $C = \dfrac{ab}{\sqrt{a^2+b^2}}$, show that the points where a tangent to S cuts the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
subtend a right angle at $O$.

Part (i) can be done by considering gradients of lines from $O$ to each of the points, multiplying them and setting equal to $-1$
Part (ii) gives the result $x\cos t+y\sin t=C$
With part (iii), I have attempted to simply rearrange for $y$ in the tangent equation, subbing into the ellipse equation and trying to solve for $x$ but this results in a huge amount of tedious algebra and I am unable to simplify it properly.
I then attempted to parameterize the ellipse in the form $x=a\cos T, y=b\sin T$ but I am unsure how to go from there to solve for the points. I did attempt to use harmonic addition and this does give some exact solutions in terms of $a,b,t$ but to simplify requires identities for $\sin(\arccos(x))$ and other similar identities like $\arctan\left(\frac{b}{a}\tan t\right)$.
Some useful information may be that we can let the points of intersections be $R(a \cos P,b\sin P)$ and $L(a\cos Q,b\sin Q)$ and substitute these points into the equation for the tangent. Now my problem is taking those equations and getting the equation we want from (i)
 [/INST] (i) We have $P = (a \cos A, b \sin A)$ and $Q = (a \cos B, b \sin B) $
Since $OP \perp OQ$ , then by using the dot product we get
$ a^2 \cos A \cos B + b^2 \sin A \sin B = 0 $
(ii) A point on the circle with radius $C$ centered at the origin is parameterized as
$ P = C (\cos t, \sin t)$
The tangent vector at $P$ is along the vector $(-\sin t , \cos t ) $ (Perpendicular to the radius vector).
Hence, the parametric equation of the tangent is
$ \ell(t,s) = (C \cos t - s \sin t , C \sin t + s \cos t ) $
(iii)  Now the above tangent line is to intersect the ellipse, and we expect two intersection points.  Let $P = (a \cos A, b \sin A) $ be the intersection point, then $P$ lies on $\ell$.
Hence
$ a \cos A = C \cos t - s \sin t \hspace{30pt} b \sin A = C \sin t + s \cos t $
Solving for $s$ from the above two equations
$ s = \dfrac{ a \cos A - C \cos t }{-\sin t} = \dfrac{ b \sin A - C \sin t }{\cos t } $
Cross-multiplying yields
$ a \cos A \cos t - C \cos^2 t = - b \sin A \sin t + C \sin^2 t $
So that
$ a \cos A \cos t + b \sin A \sin t = C \hspace{30pt}(1) $
Similarly, if $Q= (a \cos B, b \sin B)$ is the other intersection point, then
$ a \cos B \cos t + b \sin B \sin t = C \hspace{30pt}(2)$
Equations $(1)$ and $(2)$ are linear in $\cos t $ and $\sin t $.  Therefore, we can solve for them, and using Cramer's rule (for example), or any other method, we get
$ \cos t = \dfrac{ C b (\sin B - \sin A) }{ ab (\cos A \sin B - \cos B \sin A)} $
$ \sin t = \dfrac{ C a (\cos A - \cos B) }{ ab (\cos A \sin B - \cos B \sin A) } $
Now we can use the fact that $\cos^2 t + \sin^2 t = 1 $ to write
$ C^2 \bigg(b^2 (\sin B - \sin A)^2 + a^2 (\cos A - \cos B)^2 \bigg) = (ab)^2 \bigg( \cos A \sin B - \cos B \sin A \bigg)^2 \hspace{15pt}(3)$
Now, we're given that $ C = \dfrac{ab}{\sqrt{a^2 + b^2} } $
Substituting this into $(3)$,
$b^2 (\sin B - \sin A)^2 + a^2 (\cos A - \cos B)^2 = (a^2 + b^2) \bigg( \cos A \sin B - \cos B \sin A \bigg)^2 \hspace{20pt}(4)$
Expanding both the left and right sides of $(4)$,
$ b^2 (\sin^2 B + \sin^2 A - 2 \sin A \sin B) + a^2 (\cos^2 A + \cos^2 B - 2 \cos A \cos B) = (a^2 + b^2) (\cos^2 A \sin^2 B + \cos^2 B \sin^2 A - 2 \cos A \cos B \sin A \sin B ) $
$ b^2 \sin^2 B (1 - \cos^2 A) + b^2 \sin^2 A (1 - \cos^2 B) + a^2 \cos^2 A (1 - \sin^2 B) + a^2 \cos^2 B (1 - \sin^2 A) - 2 b^2 \sin A \sin B - 2 a^2 \cos A \cos B + 2 (a^2 + b^2) \cos A \cos B \sin A \sin B = 0 $
And this becomes, after using $ \cos^2 A + \sin^2 A = 1 $ and $ \cos^2 B + \sin^2 B = 1 $, and dividing through by $2$:
$ b^2 \sin^2 A \sin^2 B + a^2 \cos^2 A \cos^2 B - a^2 \cos A \cos B - b^2 \sin A \sin B + (a^2 + b^2) \cos A \cos B \sin A \sin B = 0 $
Collecting like terms,
$ (a^2 \cos A \cos B + b^2 \sin A \sin B) ( \sin A \sin B + \cos A \cos B - 1) = 0 $
But $\sin A \sin B + \cos A \cos B - 1 = \cos(A - B) - 1 \ne 0 $ (because $A \ne B$).
Therefore,
$ a^2 \cos A \cos B + b^2 \sin A \sin B = 0 $
And from part (i), this shows that the intersection points subtend a right angle at the origin.
</s>