<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] The work is indeed tedious.
Suppose that the tangent cuts the ellipse at $P(a\cos A, b\sin A)$ and $Q(a\cos B, b\sin B)$, then the equation of the chord $PQ$ is:
$$\frac{y-b\sin B}{x-a\cos B}=\frac{b(\sin B-\sin A)}{a(\cos B-\cos A)}$$
\
$$\iff ay(\cos B-\cos A) - ab\sin B(\cos B-\cos A)= bx(\sin B-\sin A)-ab(\sin B-\sin A)$$
\
$$\iff bx(\sin B-\sin A)+ay(\cos A-\cos B)=ab\sin (B-A)$$
\
$$\iff bx\left(2\sin \frac{B-A}{2}\cos \frac{A+B}{2}\right)+ay\left(2\sin \frac{B-A}{2}\sin \frac{A+B}{2}\right)=ab\left(2\sin \frac{B-A}{2}\cos \frac{B-A}{2}\right)$$
\
$$\iff bx\left(\cos \frac{A+B}{2}\right)+ay\left(\sin \frac{A+B}{2}\right)=ab\left(\cos \frac{B-A}{2}\right)$$
\
$$\iff bx\left(\frac {\cos \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}}\right)+ay\left(\frac {\sin \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}}\right)=\frac{ab}{\sqrt{a^2+b^2}}$$
Since the tangent cuts the ellipse at $P$, $Q$
This is the same as the equation of the tangent
$$x \cos t + y \sin t = \frac{ab}{\sqrt{a^2+b^2}}$$
Comparing the two equations, we have
$$\cos t = \frac {b\cos \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}}$$
and
$$\sin t = \frac {a\sin \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}}$$
Thus we have $$\left(\frac {b\cos \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}} \right)^2+\left(\frac {a\sin \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}} \right)^2=1$$
\
$$b^2\cos^2\left(\frac{A+B}{2} \right)+a^2\sin^2\left(\frac{A+B}{2} \right)=\left(a^2+b^2 \right)\cos^2 \left(\frac {B-A}{2}\right)$$
\
$$a^2\left[\cos^2\left(\frac {B-A}{2}\right)-\sin^2\left(\frac{A+B}{2} \right)\right]+b^2\left[\cos^2\left(\frac {B-A}{2}\right)-\cos^2\left(\frac{A+B}{2} \right)\right]=0$$
\
$$a^2\left[\frac{1+\cos(B-A)}{2}-\frac{1-\cos(A+B)}{2} \right]+b^2\left[\frac{1+\cos(B-A)}{2}-\frac{1+\cos(A+B)}{2}\right]=0$$
\
$$a^2\left[\cos(B-A)+\cos(A+B)\right] + b^2\left[\cos(B-A)-\cos(A+B) \right]=0$$
\
$$a^2(2\cos A \cos B)+ b^2(2\sin A \sin B)=0$$
\
$$a^2\cos A \cos B+ b^2\sin A \sin B=0$$
which is the condition for $OP$ and $OQ$ to be perpendicular.
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